Question

A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground

Answer 1

V=34.2 m/s

Step-by-step explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

`V^2_y=U^2_y+2 g h`

Now by putting the values in the above equation we got

`V^2_y=U^2_y+2 g h`

`V^2_y=0^2+2* 9.81 * 54`

Assume
`g= 9.81 m/s^2`

Thus

`V^2_y=1059.48`

`V_y=√(1059.48)\ m/s`

`V_y=32.54 m/s`

We also know that

`V_y=U_y+ g* t`

`32.54=9.81* t`

`t=(32.54)/(9.81)=3.31\ s`

In horizontal direction :

`x=U_x* t`

`U_x=(35)/(3.31)=10.54\ m/s`

Thus the resultant velocity

`V=√(V^2_y+U^2_x)`

`V=√(32.54^2+10.54^2)=34.2\ m/s`

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.