Free Fall Numericals class 9
Last updated on June 2nd, 2023 at 08:39 am
In this post named Free Fall Numericals class 9, we will solve a set of numerical problems based on the free fall topic of the class 9 science syllabus. First, we will list down the formulas of free fall and then solve the numerical problems.
Free fall Formulas
Equations of motion for freely falling bodies [downward motion]:
v =u + gt
h=ut + (1/2)gt2
v2 = u2 + 2gh
[Here, g = acceleration due to gravity]Equations of motion for freely falling bodies [upward motion]
As g becomes negative, in this case, the above equations can be used with a small but vital change.
v =u – gt
h=ut – (1/2)gt2
v2 = u2 – 2gh
[Here, g = acceleration due to gravity]
Important points to remember – for the numerical problems related to free fall
(a) When a body is falling vertically downwards, its velocity is increasing, so the acceleration due to gravity, g, is taken as positive. That is, Acceleration due to gravity = + 9.8 m/s2 for a freely falling body
(b) When a body is thrown vertically upwards, its velocity is decreasing, so the acceleration due to gravity, g, is taken as negative. That is, Acceleration due to gravity = – 9.8 m/s2 for a body thrown upwards.
(c) When a body is dropped freely from a height, its initial velocity ‘u’ is zero.
(d) When a body is thrown vertically upwards, its final velocity ‘v’ becomes zero.
(e) The time taken by a body to rise to the highest point is equal to the time it takes to fall from the same height.
Free Fall Numericals for class 9 – solved
Numerical Problem 1] To estimate the height of a bridge over a river, a stone is dropped freely into the river from the bridge. The stone takes 4 seconds to touch the water surface in the river. Calculate the height of the bridge from the water level (g = 9.8 m/s2).
Solution.
The stone is dropped freely from rest, so the initial velocity of the stone, u = 0.
As the velocity of the stone is increasing as it comes down (free fall), the acceleration due to gravity, g, is to be taken as positive.
Now, the Initial velocity of the stone, u = 0
Time taken, t = 4 s
Acceleration due to gravity, g = 9.8 m/s2 (Stone comes down)
And, Height of the bridge, h =? (To be calculated)
We know that for a freely falling body :
Height, h = ut + (1/2) gt2
Putting the above values in this formula, we get :
h = 0 × 2 + (1/2) × 9.8 × (4)2
or h =(1/2) × 9.8 × 16
or h = 78.4 m
Thus, the height of the bridge above the water level is 78.4 meters.
Numerical Problem 2] A cricket ball is dropped from a height of 40 meters.
(a) Calculate the speed of the ball when it hits the ground.
(b) Calculate the time it takes to fall through this height. (g = 10 m/s2)
Solution.
(a) Here, Initial speed, u = 0
Final speed, v=? (To be calculated)
Acceleration due to gravity, g = 10 m/s2 (Ball comes down)
And, Height, h = 40 m
Now, we know that for a freely falling body: v2 = u2 + 2gh
So, v2 = (0)2 + 2 × 10 × 40
v2 = 800
v = 8001/2 = 28.28 m/s
Thus, the speed of the cricket ball when it hits the ground will be 28.28 meters per second.
b)
Now, Initial speed, u = 0
Final speed, v = 28.28 m/s (Calculated above)
Acceleration due to gravity, g = 10 m/s2
And, Time, t =? (To be calculated)
Putting these values in the formula :
v = u + gt,
we get 28.28 = 0 + 10 × t
10 t = 28.28
t = 28.28/10 = 2.8 Sec
Thus, the ball takes 2.8 seconds to fall through a height of 40 meters.
Numerical Problem 3]. A ball is thrown up at a speed of 15 m/s. How high will it go before it begins to fall ? (g = 9.8 m/s2)
Solution.
Here the ball is going up against gravity, so the value of g is to be taken as negative (retardation)
Here, the Initial speed of the ball, u = 15 m/s
Final speed of ball, v = 0 (The ball stops)And, Height, h =? (To be calculated)
Now, putting all these values in the formula for upward motion,
v2 = u2 – 2gh,
02 = 152 – 2×9.8xh
=> h = 152 / (2×9.8) = 225/19.6 m = 11.4 m
Practice Problems (Free Fall)
Practice Problem 1] When a ball is thrown vertically upwards, it goes through a distance of 19.6 m before coming down. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, g = 9.8 m/s2)