Question

A ball is thrown vertically upwards with a velocity of $49\text{ m/s}$. Calculate
1. The maximum height of the ball.
2. The total time it takes to return to the surface of the earth.

Answer 1

Hint: When a ball is thrown upwards with a certain velocity, the ball moves against the acceleration due to gravity and attains a maximum height. At the point of maximum height, the velocity of the ball is zero. After that the ball is accelerated downwards by gravity and the time taken for the ball to cover the maximum height is downwards is noted. The total time will be the sum of the time taken by the ball to reach the maximum height and the time taken to return to the ground.
Formula Used:
The distance covered by a body thrown vertically upwards with a velocity u is given by,
$s=\dfrac{{{u}^{2}}}{2g}$
The velocity of a body which has an initial velocity u and moving with an acceleration a at a time t is given by,
$v=u+at$

Complete step by step answer:
The ball is projected upwards with a velocity of $49\text{ m/s}$, the maximum it can reach can be found out using the formula,
$s=\dfrac{{{u}^{2}}}{2g}$
Where
s is the maximum height.
u is the initial velocity.
g is the acceleration due to gravity.
Substituting the values in the above equation we get,
$s=\dfrac{{{\left( 49m{{s}^{-1}} \right)}^{2}}}{2\times 9.8m{{s}^{-2}}}$
$\therefore s=122.5m$ … equation (1)
So, the maximum height gained by the ball is 122.5m.
The time taken to reach this height can be found out using the formula,
$v=u+at$
Where
u is the initial velocity.
v is the final velocity.
a is the acceleration of the body.
In our case the final velocity is zero and a=-g. So, we can write,
$t=\dfrac{u}{g}=\dfrac{49m{{s}^{-1}}}{9.8m{{s}^{-2}}}$
$\therefore t=5\text{ seconds}$ … equation (2)
So, the time taken to reach the maximum height is 5 seconds.
The time taken by the ball to come back to the ground from maximum height can be found out using,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
In our case the initial velocity can be considered zero, a=g. So. We can write,
$s=\dfrac{1}{2}g{{t}^{2}}$
s is the maximum height achieved by the ball.
$122.5m=\dfrac{1}{2}\times 9.8m{{s}^{-2}}\times {{t}^{2}}$
$\therefore t=5\text{ seconds}$ … equation (3)
So, the total time taken by the ball to reach the ground is the sum of time taken to reach the maximum height and the time taken to reach the earth from the maximum height.
$T=5s+5s$
$\therefore T=10s$
So, the total time taken is 10 seconds.

Note:
A body thrown upwards can be considered as a projectile projection with velocity in the horizontal direction as zero.
A body is said to be under freefall if the body is only falling under the gravity of the earth. A body under freefall will feel complete weightlessness, provided air friction and other dissipative forces are neglected.