a stone is thrown vertically upward with a speed of 29.4 m/s. find -

1. the time taken by the stone to reach the maximum height.

2. maximum height reached by the body and

3. show that time of ascent is equal to time of descent.

We have,

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

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