# A ball is thrown vertically upwards with a velocity of 49 ms–1.

Question.
A ball is thrown vertically upwards with a velocity of 49 ms–1. Calculate
(i) The maximum height to which it rises.
(ii) The total time it takes to return to the surface of the earth.

Solution:

(i) Initial velocity of the ball, $\mathrm{u}=49 \mathrm{~ms}^{-1}$

Final velocity of the ball, $\mathrm{v}=0$

Acceleration due to gravity $g=-9.8 \mathrm{~ms}^{-2}$

[In upward direction, $g$ is taken -ve]

Height attained by the ball, $\mathrm{s}=$ ?

Time for rising up, t = ?

We know, $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{gs}$

$(0)^{2}-(49)^{2}=2 \times(-9.8) \times \mathrm{s}$

$s=\frac{-49 \times 49}{-2 \times 9.8}=122.5 \mathrm{~m}$

Aliter : Maximum height achieved is given by,

$\mathrm{H}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{(49)^{2}}{2 \times 9.8}=\frac{49 \times 49}{2 \times 9.8}=122.5 \mathrm{~m}$

(ii) We know $\mathrm{v}=\mathrm{u}+\mathrm{gt}$

$0=49-9.8 \times t$

$\Rightarrow \mathrm{t}=\frac{49}{9.8}$

$\Rightarrow \mathrm{t}=5 \mathrm{~s}$

Now, time for upward journey of the ball = the time for downward journey of the ball.

$\therefore$ Total time taken by the ball to return to the surface of earth $=2 \times t=2 \times 5=10 \mathrm{~s}$

[Aliter:Total time of journey, $T=\frac{2 \mathrm{u}}{\mathrm{g}}=\frac{2 \times 49}{9.8}=10 \mathrm{~s}$ ]