Question

A stone is thrown vertically upward with a speed of 29.5 m/s .
Find -
1. The time taken by the stone to reach the maximum height.
2. Maximum height reached by the body.
3. Show that time of ascend = time of descend .

Answer 1

Initial velocity, u = 29.4 m/s (upward)

Acceleration due to gravity, g = -9.8 m/s (downward)

Using,

v = u + at

=> 0 = 29.4 – 9.8t

=> t = 3 s

Using,

v2 = u2 + 2as

=> 0 = u2 - 2gh

=> h = u2/2g = 44.1 m

When it falls down from the maximum height, its initial velocity is zero.

Using,

S = ut + ½ at2

=> -44.1 = 0 (1/2)(-9.8)t2  [negative sign with height indicates that the displacement is now downward]

=> t = 3 s

Time of ascent = time of descent = 3 s