Question

A ball is thrown vertically upward . It has a speed of 10ms^-1 when it has reached one half of its maximum height . How high does the ball rise ? Take g = 10 ms^-1

Answer 1

Answer: 10 metres


Let us suppose that the ball was thrown with an initial speed $u$
Now, let us assume the Maximum Height reached is $H$

A constant acceleration $g = 10 \, \, m/s^2$ is acting in downward direction. 


Let us consider the motion from start to Max Height. In this time interval, final velocity is zero, and distance travelled is H. We have:

$v^2 = u^2 + 2as \\ \\ \\ \implies 0^2 = u^2 + 2(-10)(H) \\ \\ \\ \implies u^2 = 20H$

Now, let us consider the time from start to when the ball reaches half the maximum height. Now, in this time interval, final velocity is 10 m/s, and Distance Travelled is H/2

$v^2=u^2+2as \\ \\ \\ \implies 10^2 = 20H + 2(-10)\left( \frac{H}{2} \right) \\ \\ \\ \implies 100 = 20H - 10H \\ \\ \\ \implies 10H = 100 \\ \\ \\ \implies \boxed{H = 10 \, \, metres}$

Thus, maximum height reached is 10 metres.

Answer 2

Answer:

10

Explanation:

Time is 1 second for zero velocity i.e max height.

V=u+at

Put v zero,u 10 and a=g

So a/c to question velocity is 10m/s

Hence in 1s distance travelled 10m.