A ball is thrown vertically upwards with a velocity of 49m/s.
Calculate
(a)the maximum height to which it rises.
(b)the total timeit takes to return to the surface of the earth.

:confused:

(X^2 = square of x)

Use the formula: v^2 - u^2 = 2as

Here v = final velocity = 0, u = initial velocity = 49, a = acceleration upward = -g

0 - 49^2 = -2gs

2gs = 49^2

Use usual value of g

So s = distance = 49^2/2g

This is the maximum height = s


For downward time:

Initial velocity u = 0, acceleration a = g, distance = s (found above)

Formula: s = ut + 1/2 at^2 will give you the time t
s = 1/2 gt^2, t^2 = 2s/g =...