A ball is thrown horizontally from a height of $\mathbf{20}\mathbf{m}$and hits the ground with a speed that is three times its initial speed. What is the initial speed?

1) The final speed of a ball is three times its initial speed.

2) Height above the ground is, $\mathrm{h}=20\mathrm{m}$.

Using the equations of projectile motion, we can find the initial velocity of a ball.When the object is thrown in the horizontal direction, its vertical component of the velocity is zero. As it moves down, it gains speed. Using this information and the kinematic equations, we can calculate the initial speed.

Formula:

$v^2=V_0^2+2a\left(y-y_0\right)$ …(i)

Let the initial velocity of a ball is$\mathrm{Vm}/\mathrm{s}$.

So, the horizontal velocity, of a ball is${\mathrm{V}}_{\mathrm{x}}=\mathrm{Vm}/\mathrm{s}$.

The initial vertical velocity of the ball is zero.

When the ball strikes the ground, it will have horizontal as well as vertical velocity. Let us assume that its vertical velocity when the ball is about to hit the ground is${\mathrm{V}}_{\mathrm{y}}\mathrm{m}/\mathrm{s}$.

So, the final speed $3\mathrm{V}$is the magnitude of the final velocity. Hence,

$$\begin{gathered} 3\mathrm{V}=\sqrt{{\mathrm{V}}_{\mathrm{x}}^2+{\mathrm{V}}_{\mathrm{y}}^2} \\ {\left(3\mathrm{V}\right)}^2={\mathrm{V}}^2+{\mathrm{V}}_{\mathrm{y}}^2 \\ {\mathrm{V}}_{\mathrm{y}}^2=8{\mathrm{V}}^2 \end{gathered}$$

Now, use the kinematic equation for the final velocity in the vertical direction.

From equation (i), we can write that

$\begin{array}{l}{\mathrm{V}}_{\mathrm{y}}^2={\mathrm{v}}_0^2-2\mathrm{g}(\mathrm{y}-{\mathrm{y}}_0)\\ 8{\mathrm{V}}^2=0^2-2\mathrm{x}\times (9.8\mathrm{m}/\mathrm{s})\times (0\mathrm{m}-20\mathrm{m})\\ \mathrm{V}=7.0\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the initial velocity with which the ball was thrown is $7.0\mathrm{m}/\mathrm{s}$.