Question

A stone thrown vertically upwards takes 3 seconds to attain maximum height, calculate

(i) initial velocity of stone

(ii) maximum height attained by the stone

Answer 1

Let,

u = Initial velocity of stone

H = Maximum height attained by the stone

v = final velocity at maximum height = 0

t = time taken to attain maximum height = 3 s

a = -g = acceleration due to gravity

(i) We know that,

v = u + at

0 = u – g × 3

u = 3g = 3 × 9.8 ms-2

u = 29.4 ms-1

(ii) From the equations of motion, we know that,

S = ut + \(\frac{1}{2}\)at²

H = 29.4 x 3 + \(\frac{1}{2}\)9.8 x 9

H = 88.2 + 44.1

H = 132.3 m