A stunt man drives a car at a speed of 20 m/s off a 30-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20°.

a. How far from the base of the cliff does the car land?

b. What is the car’s impact speed?

The initial speed of the stunt man is$20\text{m/s}$, the height of the cliff is$30\text{m}$and the angle of inclination is$20°$.

The formula to calculate the vertical component of the initial speed of the stunt man is given by

$v_y=v_0\sin \theta ........................\left(1\right)$

Here, $v_y,v_0,\theta$are the vertical component of the initial speed $v_0$and $\theta$is the angle of inclination.

The formula to calculate the vertical height travelled by the stunt man is given by

localid="1648347928517" $-h=v_{y}t-\frac{1}{2}g{t}^2..........................\left(2\right)$

Here, $h,t,g$are the vertical height, the time taken and the acceleration due to gravity.

Substitute $v_0\sin \theta$for $v_y$into equation (2) and simplify to obtain the expression for the time taken by the stunt man.

localid="1648348245875" $\begin{array}{rcl}-h& =& v_0\sin \theta t-\frac{1}{2}g{t}^2\\ g{t}^2-2v_0\sin \theta t-2h& =& 0\\ t& =& \frac{2v_0\sin \theta \pm \sqrt{{\left(2v_0\sin \theta \right)}^2+4g\left(2h\right)}}{2g}.................................\left(3\right)\end{array}$

Substitute $20\text{m/s}$for $v_0$, $20°$for $\theta$, $30\text{m}$for $h$and $9.80{\text{m/s}}^2$for $g$into equation (3) and solve to calculate the value of the time taken by the stunt man.

$\begin{array}{rcl}t& =& \frac{2\times 20\times \sin 20°\pm \sqrt{{\left(2\times 20\times \sin 20°\right)}^2+4\times 9.80\times \left(2\times 30\right)}}{2\times 9.80}\\ & \approx & 3.27\text{s}\text{and}-1.87\text{s}\end{array}$

Hence, the allowed value of time is$3.27\text{s}$.

The formula to calculate the horizontal distance$\left(x\right)$ travelled by the stunt man is given by

$x=v_0\cos \theta t............................\left(4\right)$

Substitute proper values of the known parameters into equation (4) to calculate the required horizontal distance travelled by the stunt man.

$\begin{array}{rcl}x& =& 20\text{m/s}\times \cos 20°\times 3.27\text{s}\\ & \approx & 61.45\text{m}\end{array}$

Let us assume that $m,v$are the mass and the final impact speed of the car.

Hence, from energy conservation law, it can be written that

role="math" $\frac{1}{2}m{v_0}^2+mgh=\frac{1}{2}m{v}^2....................\left(5\right)$

Simplify equation (5) to obtain the impact speed of the car.

role="math" $\begin{array}{rcl}\frac{1}{2}m{v_0}^2+mgh& =& \frac{1}{2}m{v}^2\\ v^2& =& {v_0}^2+2gh\\ v& =& \sqrt{{v_0}^2+2gh}...................\left(6\right)\end{array}$

Substitute the proper values of the parameters into equation (6) to calculate the required impact speed of the car.

$\begin{array}{rcl}v& =& \sqrt{{\left(20\text{m/s}\right)}^2+2\times 9.80{\text{m/s}}^2\times 30\text{m}}\\ & \approx & 31.43\text{m/s}\end{array}$