Question

A ball is thrown vertically upwards from the ground. It crosses a point at the height of $25m$ twice at an interval of $4 sec$. The ball was thrown with the velocity of
A. $20m/s$
B. $25m/s$
C. $30m/s$
D. $35m/s$

Answer 1

To solve this problem, you can recall the concept of 2-D vertical motion. In such cases, time taken for any of the moving objects during their upward motion is exactly the same as that for the downward motion. In order to find the velocity we require maximum height so first of all we will find the maximum height with the help of time interval.

Complete step by step answer:
consider ‘A’ point at ground .From point A one ball is thrown vertically upwards to point ‘B’. From point ‘A’ to ‘B’ height is 25m after reaching at point ‘A’ ball crosses this point twice at an interval of 4sec. It means ball covered point ‘B’ to ‘C’ at time 2sec and then return back at same point with time 2sec. Let us explain with the help of the following diagram :

Let velocity at point ‘A’ is $u$. Velocity at point ‘B’ is $v$. Velocity at point ‘C’ is ${v_1}$.
Way to solve this problem:
Firstly find the height from point ‘B’ to ‘C’
We know that,
$h = \dfrac{1}{2}g{t^2}$ (Velocity at point C is 0)
$\Rightarrow g = 9.8m/s$ (i.e. nearly equal to 10m/s)
$\Rightarrow t = 2\sec$

Put the values of g and t we get the value of h
$h = \dfrac{1}{2}\times 10 \times {2^2}$
\Rightarrow h =20m
Total height from point ‘A’ to ‘C’ is
${h_{\max }} = {h_{AB}} + {h_{BC}}\\\ \Rightarrow{h_{\max }} = 25+20\\\ \Rightarrow{h_{\max }} = 45m$
Velocity is found from:
${V^2} = {u^2} + 2gh$
Here, V is velocity at point ‘C’, U is velocity at point ‘A’,g is gravity, h is height from point ‘A’ to
’C’.

Velocity at point ‘C’ is zero
${u^2} = 2gh \\\ \Rightarrow{u^2} = 2\times 10 \times 45\\\ \Rightarrow u = \sqrt {900} \\\ \therefore u =30m/s$

The ball was thrown with the velocity of 30 m/s, it means Option C is the correct answer.

Note: At the maximum height the final velocity of the body is zero. For example in the present case, the ball was thrown with the velocity of 30 m/s upwards and its velocity decreases to 20 m/s at height of 25 m from ground and at height h, ball stops moving and its velocity reduces to zero while again velocity increases to 20 m/s at a height of 25 m and it becomes 30 m/s on reaching the ground. Always remember that for a body projected vertically upwards a = -g.