Question

Freely falling body has velocity 30m/s at a point. What will be its velocity after

2sec? (g = 10m/s2

)

32m/s2 b) 60m/s2 c) 50m/s2 d) 45m/s2​

Answer 1

Answer:

Initial velocity velocity = 30 m/s upwards ( taken as positive).

Acceleration due to gravity= - 10 m/s²

Time=1 s

Distance covered in 1 second can be obtained by using the relation;

s = u t + ½ a t²;

s = 30 m/s × 1 s - ½ × 10 m/s² × 1s² = 30 m - 5 m= 25 m.

So in its last second the body while returning to earth will again cover the same distance of 25 m in its downwards journey.

Let us check.

Total time for upwards journey can be obtained using the equation;

v = u + a t,

The body shall travel upwards till its velocity becomes zero starting with initial velocity u = 30 m/s, with acceleration due to gravity= -10 m/s² reducing it. Using the equation,

0 m/s = 30 m/s - 10 m/s² × t s, or

t = 3 s.

The body takes 3 s to go up its highest point and 3 s to come down from the highest point reached. The highest point h reached can be obtained using

v² - u² = 2 g h,

Now v= 0 m/s, u = 30 m/s² , g = 10 m/s² , we get

h =-30²/- 2×10= 45 m.

Time then to free fall a height of 45 m can be obtained using the relation

s = u t + ½ a t².

In this case s= 45 m, u = 0 m/s, a + 10 m/s². On substituting we get

45 = 0×t + ½ ×10× t² , or

45= 5 t² , or

t = 3 s.

So the body takes 3 s to come down from 45 m height with initial velocity as 0 m/s, and under acceleration of gravity= + 10 m/s². The distance covered in the nth second by a body moving with initial velocity u, and uniform acceleration a is given by

X ( nth) = u×1 s + ½ a ( 2n-1)×1s

In our case u= 0 m/s, n= 3rd second, a=+10 m/s²

So distance covered in 3rd second= 0 + ½×10×(2×3–1) = 0 + 5×5= 25 m .

So distance covered in last second is 25 m.