Question

A stone is dropped from the window of a bus moving at 60km /h. If the window is 1.96 m high, find the distance along the track, which the stone moves before striking the ground.

Answer 1

Since, the stone is dropped, its initial horizontal velocity is equal to the velocity og the bus which is, ux = 60 km/h = 16.67 m/s Its initial vertical velocity is, u y = 0 The height through which the stone falls is, h = 1.96 m Using, S = ut + ½ at 2 => 1.96 = 0 + ½ (9.8)(t 2 ) => t = 0.63 s Horizontal distance moved by the stone in this time is, X = u x t = (16.67)(0.63) = 10.5 m The path of the stone can be found as follows: X = u x t = 16.67t Y = u y t + ½ at 2 = (1/2)(9.8)t 2 = 4.9t 2 Using the above two equations the position of the stone at any instant can be determined.

Answer 2

Let's get our units consistent.

Acceleration due to gravity:

g=9.8ms^2

Speed: 60kmhr^-1 =16.67ms

Distance to fall: 196cm=1cm:0.01m=1.96m

We need the time to hit the ground, so we need the formula

y=u⋅t+1/2⋅a⋅t^2

1.96m=0+1/2⋅9.8ms^2⋅t^2

t^2=2⋅1.96m9.8ms^2=0.4s^2

t=0.632s

Now we can calculate the distance down the road it travels in that time. It is

d=16.67ms⋅0.632s=10.5m

I hope this helps,