Question

A ball thrown up is caught bya thrower after 4seconds. How high did it go and with what velocity was it throwm? How far below its highest point was it 3seconds after start?

Answer 1

Given,

Time of flight of the ball is 4 s

So, it takes 2 s to reach the maximum height.

Using,

v = u + at

=> 0 = u -(9.8)(2)

=> u = 19.6 m/s

Again,

v² = u² + 2as

=> 0 = 19.6² - 2(9.8)(h)

=> h = 19.6 m

The distance travelled by the ball in 3 s is,

S = ut + 1/2 at²

=> S = (19.6)(3) - (1/2)(9.8)(3²)

=> S = 14.7 m

The distance from the maximum height is = h – S = 19.6 - 14.7 = 4.9 m