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

Answer:

V=34.2 m/s

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

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

$V^2_y=0^2+2\times 9.81 \times 54$

Assume $g= 9.81 m/s^2$

Thus

$V^2_y=1059.48$

$V_y=\sqrt{1059.48}\ m/s$

$V_y=32.54 m/s$

We also know that

$V_y=U_y+ g\times t$

$32.54=9.81\times t$

$t=\dfrac{32.54}{9.81}=3.31\ s$

In horizontal direction :

$x=U_x\times t$

$U_x=\dfrac{35}{3.31}=10.54\ m/s$

Thus the resultant velocity

$V=\sqrt{V^2_y+U^2_x}$

$V=\sqrt{32.54^2+10.54^2}=34.2\ m/s$

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.