Question

A ball is thrown vertically upwards and it returns to the thrower after 8 seconds.
Calculate
(i) velocity with which it was thrown
(ii) maximum height it acquired

Answer 1

Answer:

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be u

(a). For upward motion,

$v = u + at \\ \\ 0 = u + ( - 10) \times 3 \\ = > u = 30m/s$

(b). The maximum height reached by the ball

$h = ut + \frac{1}{2} at {}^{2} \\ h = 30 \times 3 + \frac{1}{2} ( - 10) \times 3 {}^{2} \\ h = 45m$

(c). After 3 second, it starts to fall down.

Let the distance by which it fall in 1 s be d

$d = 0 + \frac{1}{2} at {}^{2} \: \: \: where t = 15 \\ d = \frac{1}{2} \times 10 \times (1) {}^{2} = 5m$

∴ Its height above the ground, h

=45−5=40 m

Hence after 4 s, the ball is at a height of 40 m above the ground.