Free Fall Exercises

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Test your knowledge of free fall movement with the 10 questions Next. Check out the comments after the feedback to get your questions answered.

For calculations, use the formulas:

Free fall velocity: v = g.t
Height in free fall: h = gt2/2
Torricelli Equation: v2 = 2.g.h

question 1

Review the following sentences about free fall motion and judge as true (V) or false (F).

I. The mass of a body influences the free fall movement.
II. The speed of a free-falling body is inversely proportional to the duration of the movement.
III. Local gravity acceleration acts on bodies in free fall.
IV. In a vacuum, a feather and a golf ball fall with the same free fall speed.

The correct sequence is:

a) V, F, F, V
b) F, V, F, F
c) F, F, V, V
d) V, F, V, f

Correct alternative: c) F, F, V, V.

I. FALSE. Free fall is influenced by the acceleration of local gravity and, therefore, bodies with different masses would reach the ground at the same time, ignoring the friction force of the air.

II. FALSE. Speed ​​is directly proportional, as in free fall it increases at a constant rate. Note the formula below.

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V = g.t

Compare the fall time of two bodies, C1 and C2, with speeds of 20 m/s and 30 m/s, respectively:

straight V with straight C with 1 subscript end of subscript space equal to straight space g. straight t space 20 straight space m divided by straight s space equal to space 10 straight space m divided by straight s squared space. straight space t straight space t space equal to numerator space 20 straight space m divided by straight s over denominator 10 straight space m divided by straight s squared end of fraction straight t space equal to space 2 straight space s
straight V with straight C with 2 subscript subscript end of subscript space equal to straight space g. straight t space 30 straight space m divided by straight s space equal to space 10 straight space m divided by straight s squared space. straight space t straight space t space equal to numerator space 30 straight space m divided by straight s over denominator 10 straight space m divided by straight s squared end of fraction straight t space equal to space 3 straight space s

III. REAL. In free fall, the force of gravity acts on the bodies, which are not subject to other forces such as drag or friction.

IV. In this case, the only force acting on them is the acceleration of gravity, as they are under the influence of the same force then they will arrive at the same time.

question 2

Regarding the free fall movement, it is INCORRECT to state that:

a) Graphically, the variation of velocity in relation to time is an ascending straight line.
b) The free fall movement is uniformly varied.
c) The trajectory of a body in free fall is straight, vertical and oriented downwards.
d) The body in free fall presents acceleration that increases at a constant rate.

Incorrect alternative: d) The body in free fall has acceleration varying at a constant rate.

In free fall motion, acceleration is constant, which increases at a constant rate is velocity.

Because it is a uniformly varied movement, the velocity vs. time graph of the free fall movement is an ascending straight.

The initial velocity in the free fall movement is nil. When the body is abandoned, it follows a straight, vertical and downward-oriented trajectory.

question 3

Under gravity acceleration of 10 m/s2, what is the speed that a drop of water falls from the tap at a height of 5 m, considering that it started from rest and the air resistance is nil?

a) 5 m/s
b) 1 m/s
c) 15 m/s
d) 10 m/s

Correct alternative: d) 10 m/s

For this question we will use the formula the Torricelli equation.

straight v squared space equals space 2. straight g. straight h space straight space v squared space equal to space 2.10 straight space m divided by straight s squared space. space 5 straight space m straight space v squared space equal to space 100 straight space m squared divided by straight s squared straight v space equal to root square of space 100 straight space m squared divided by straight s squared end of root straight space v space equal to space 10 straight space m divided by straight only

Therefore, a drop starting from 5 meters high acquires a velocity of 10 m/s.

question 4

How long, approximately, does a fruit that fell from a tree, located at a height of 25 m, take to reach the ground? Neglect the air resistance and consider g = 10 m/s2.

a) 2.24 s
b) 3.0 s
c) 4.45 s
d) 5.0 s

Correct alternative: a) 2.24 s.

For this question we will use the free fall height formula.

straight h space equal to space gt squared over 2 space space double right arrow t square space equal to space numerator 2. straight h over straight denominator g end of fraction numerator 2.25 straight space m over denominator 10 straight space m divided by straight s squared end of fraction space equal to straight space t squared space space 50 straight space m divided by 10 straight space m divided by straight s squared space space equal to straight space t squared straight t space equal to space square root of 5 straight space s squared end of root space straight space t space equal to space 2 comma 24 space straight only

Thus, the fruit falling from the tree will touch the ground after 2.24 seconds.

question 5

Neglecting air resistance, if a vase that was on a balcony fell, taking 2 seconds to reach the ground, how high was the object? Consider g = 10 m/s2.

a) 10 m
b) 20 m
c) 30 m
d) 40 m

Correct alternative: b) 20 m.

To determine how high the object was, we will use the following formula.

straight h space equal to space gt squared over 2 space space space space straight h space equal to space numerator 10 space. space 2 squared over denominator 2 end of fraction straight h space equal to space numerator 10.4 over denominator 2 end of fraction straight h equals space 40 over 2 straight h space equals space 20 straight space m

Therefore, the object was at a height of 20 meters and when it fell it hit the ground in 2 seconds.

question 6

A bowling ball was dropped from a balcony 80 meters above the ground and acquired a free-fall movement. How high was the ball after 2 seconds?

a) 60 m
b) 40 m
c) 20 m
d) 10 m

Correct alternative: a) 60 m.

Using the hourly space equation we can calculate the position of the bowling ball in a time of 2 seconds.

straight S space equal to space 1 half gt squared straight space S space equal to space 1 half 10 straight space m divided by straight s squared. space left parenthesis 2 straight s right parenthesis squared straight S space equals space 5 straight space m divided by straight s squared space. space 4 straight space s squared straight S space equal to space 20 straight space m

Next, we subtract the total height from the previously calculated height.

h = 80 - 20 m
h = 60 m

Thus, the bowling ball was at 60 meters after 2 seconds of the beginning of the movement.

question 7

(UFB) Two people find themselves falling from the same height, one with the parachute open and the other with it closed. Who will reach the ground first, if the means is:

a) the vacuum?
b) the air?

Right answer:

a) In a vacuum, both people will arrive at the same time, as the only force that will act on them is gravity.

b) With air resistance, the person with the open parachute will be more influenced, as it causes a delaying effect on movement. In this case, the latter will arrive first.

question 8

(Vunesp) A body A is dropped from a height of 80 m at the same instant that a body B is launched vertically downwards with an initial velocity of 10 m/s from a height of 120 m. Neglecting air resistance and considering gravity acceleration to be 10 m/s2, it is correct to state, about the movement of these two bodies, that:

a) Both reach the ground at the same moment.
b) Body B reaches the ground 2.0 s before body A
c) The time taken for body A to reach the ground is 2.0 s less than the time taken by B
d) Body A hits the ground 4.0 s before body B
e) Body B hits the ground 4.0 s before body A

Correct alternative: a) Both reach the ground at the same instant.

Let's start by calculating the time of body A.

straight h space equal to space 1 half gt squared space 80 straight space m space equal to space 1 half gt squared space 80 straight space m space equal to space 1 half 10 straight space m divided by straight s squared straight t squared space 80 straight space m space equal to space 5 straight space m divided by straight s squared straight t ao square straight space t squared space equal to numerator space 80 straight space m over denominator 5 straight space m divided by straight s squared end of fraction straight t ao square space equal to 16 straight space s squared straight t space equal to space square root of 16 straight space s squared end of straight root t space equal to space 4 straight space s

Now we calculate the time of body B.

straight h space equal to straight space v with 0 subscript straight t space plus space 1 half gt squared 120 straight space m space equal to space 10 straight space m divided by straight s. straight t space plus 1 half 10 straight space m divided by straight s squared straight t squared space 120 space equal to 10 space. straight t space plus space 5 straight t squared 5 straight t squared space plus space 10 straight t space minus space 120 space equal to space 0 space left parenthesis divided by 5 right parenthesis straight t squared space plus space 2 straight t space minus space 24 space equal to space 0

As we arrive at an equation of the 2nd degree, we will use Bhaskara's formula to find the time.

numerator minus space b space plus or minus space square root of b squared space minus space 4 a c end of root over denominator 2 end of fraction numerator minus space 2 space plus or minus space square root of 2 squared space minus space 4.1. left parenthesis minus 24 right parenthesis end of root over denominator 2.1 end of fraction numerator minus space 2 plus or minus space square root of 4 space plus space 96 end of root over denominator 2 end of fraction numerator minus space 2 plus or minus space square root of 100 over denominator 2 end of fraction numerator minus space 2 plus or minus space 10 over denominator 2 end of fraction double right arrow table row with cell with t apostrophe space equal to space numerator minus space 2 space plus space 10 over denominator 2 end of fraction equal to 8 over 2 equals 4 space end of cell row with cell with t apostrophe apostrophe space equal to space numerator minus space 2 space minus 10 space over denominator 2 end of fraction equals numerator minus 12 over denominator 2 end of fraction equals minus 6 end of cell end from the table

As time cannot be negative, the time of body b was 4 seconds, which is equal to the time that the body A took and, therefore, the first alternative is correct: the two reach the ground in the same instant.

question 9

(Mackenzie-SP) Joãozinho leaves a body at rest from the top of a tower. During free fall, with constant g, he observes that in the first two seconds the body travels distance D. The distance traveled by the body in the following 4 s will be:

a) 4D
b) 5D
c) 6D
d) 8D
e) 9D

Correct alternative: d) 8D.

The distance D in the first two seconds is given by:

straight D space equals space 1 half gt squared straight D space equals space 1 half straight g.2 squared straight D space equal to space numerator 4 straight g over denominator 2 end of fraction straight D space equal to space 2 straight g

The distance in the next 4 seconds indicates that the body was already in 6 seconds of movement and, therefore:

straight H space equals space 1 half gt squared straight H space equals space 1 half straight g.6 squared straight H space equal to space numerator 36 straight g over denominator 2 end of fraction straight H space equal to space 18 straight g

Thus, the distance in 4 seconds is the difference between H and D.

H - D = 18g - 2g
H - D = 16g

As the answer must be given in terms of D, as D represents 2g, then 16g is the same as 8D and so alternative d is correct.

question 10

(UFRJ) A body in free fall travels a certain vertical distance in 2s; therefore, the distance traveled 6 s will be

a) double
b) triple
c) six times larger
d) nine times larger
e) twelve times larger

Correct alternative: d) nine times larger.

1st step calculate the vertical distance in 2s time.

straight S with 1 subscript space equal to space 1 half gt squared straight S with 1 subscript space equal to space 1 half 10.2 squared straight S with 1 subscript space equal to space 40 over 2 straight S with 1 subscript space equal to space 20 space straight m

2nd step: calculate the distance covered in 6 s.

straight S with 2 subscript space equal to space 1 half gt squared straight S with 2 subscript space equal to space 1 half 10.6 ao square straight S with 2 subscript space equal to space 360 ​​over 2 straight S with 2 subscript space equal to space 180 straight space m

180 m is 9 times larger than 20 m. Therefore, alternative d is correct.

Get more knowledge by reading the texts below.

  • Free fall
  • gravitational force
  • Vertical launch
  • gravity acceleration
  • evenly varied movement
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