Question

A boy starts at rest and slides down a frictionless slide as in Figure P8.45. The bottom of the track is a height h above the ground. The boy then leaves the track horizontally, striking the ground at a distance d as shown. Using energy methods, determine the initial height H of the boy above the ground in terms of h and d.

Short answer

The initial height H of the boy above the ground in terms of h and d,

$H=h+\frac{D^2}{4h}$

Step-by-step solution

Information Given

The energy stored in an object by virtue of its position is called potential energy.

$\mathit{U}\mathbf{=}\mathit{m}\mathit{g}\mathit{R}$

The energy stored in an object by virtue of its motion is called potential energy.

role="math" localid="1663585810859" $\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Here,

U= Potential energy

K=Kinetic energy

v= Speed of the body

g= Gravitational acceleration

R= Displacement in meter

m= Mass of the body

Calculation

Let $y=0$at ground level. The boy starts from rest $\left(v_i=0\right)$at the top of the slide $y_i=H$. The instant he leaves the lower end $\left(y_f=h\right)$of the frictionless slide at speed v, where his velocity is horizontal$\left(v_{xf}=v,v_{yf}=0\right)$. By using the concept from step (1), we have

$\begin{array}{l}{E}_0=E_{top}\\ \frac{1}{2}m{v}^2+mgh=0+mgH\\ v^2=2g\left(H-h\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\end{array}$

Consider his flight as a projectile after leaving the end of the slide, so for the vertical motion,

$\Delta y=V_{yi}t+\frac{1}{2}{a}_y{t}^2$

The time to reach the ground is as follows:

$\begin{array}{c}-h=0+\frac{1}{2}\left(-g\right)t^2\\ t=\sqrt{\frac{2h}{g}}\end{array}$

The horizontal distance traveled (at constant horizontal velocity) during this time is d,

$\begin{array}{c}d=-vt\\ =v\sqrt{\frac{2h}{g}}\\ v=d\sqrt{\frac{g}{2h}}\\ =\sqrt{\frac{g{d}^2}{2h}}\end{array}$

From the above values and equation (1),

$\begin{array}{l}\frac{g{d}^2}{2h}=2g\left(H-h\right)\\ H=h+\frac{d^2}{4h}\end{array}$

The initial height of the boy is given as

$H=h+\frac{D^2}{4h}$