Question

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

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

Answer 1

Given initial velocity u = 49 m/s
$(i) \: at \: the \: maximum \: height \: velocity \: becomes \: zero \\ \\ therefore \: final \: velocity \: \: v = 0 \\ \\ from \: the \: third \: equation \: of \: upward \: motion \: \\ \\ v {}^{2} = u {}^{2} - 2gh \\ \\ 0 = (49) {}^{2} - 2 \times 9.8 \times h \\ \\ h = \frac{49 \times 49}{2 \times 9.8} = 122.5m \\ \\ therefore \: maximum \: height \: attained = 122.5m \\ \\ \\ (ii) \: from \: the \: first \: equation \: of \: motion \\ v = u - gt \\ \\ or \: 0 = 49 - 9.8 \times t \\ \\ t = \frac{49}{9.8} = 5 \: second$
For the motion against gravity, the time of descent is same as the time of ascent.
So time taken by the ball to fall from maximum height is 5 second.

$therefore \: total \: time \: taken \: by \: the \: ball \: to \: return \: to \: surface \: of \: the \: earth \\ = 5 + 5 \\ = 10 \: second \\ \\ hope \: it \: helps$

Answer 2

I hope my answer is right