Question

A ball thrown up vertically returns to the thrower
after 6 s. Find :
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.​

Answer 1

Explanation:

For upward motion,

v=u+at

∴ 0=u+(−10)×3

⟹u=30 m/s

(b). The maximum height reached by the ball

h=ut+½at²

h=30+ ½(-10)×3²

h=45 m

(c). After 3 second, it starts to fall down.

Let the distance by which it fall in 1 s be d

d= ut+½t². t'=1s. ut=0

=5 m

∴ Its height above the ground, h

′

=45−5=40 m

Hence after 4 s, the ball is at a height of 40 m above the ground.

hope it helps ☺️