Question

A body is thrown vertically upward from top of a building which is 30 m tall. The body goes 5 m high above the top of the building and then falls towards the ground and finally hits the ground. Find its distance and displacement.

Answer 1

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at

2

=20(2)+0.5(−10)(2)

2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at

2

45=0+0.5(10)(t

′

)

2

t

′

=3s

Total time= 3+2= 5s

Answer 2

Explanation:

Distance= 30+5= 35m

Displacement= 0+30= 30m

Displacement is zero because the body first goes up and then returns to the initial position from where it was thrown. Then it further goes down and covers a distance of 30m.

Hope it will help you