Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.

Answer 1

(i) According to the third equation of motion under gravity:
$v^2-u^2=2gs$
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s
During upward motion, $g=-9.8m{s}^{-2}$
Let h be the maximum height attained by the ball.
Hence,
$0-{49}^2=2\times -9.8\times h$
$h=\frac{49\times 49}{2\times 9.8}=122.5m$

(ii) Let t be the time taken by the ball to reach the height 122.5 m, then according to the first equation of motion:
v = u + gt
We get,
$0=49+t\times (-9.8)$
$9.8t=49$
$t=\frac{49}{9.8}=5s$
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 s + 5 s = 10 s