Question

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3–41). (a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed?

FIGURE 3-41Problem 37

Short answer

(a) The minimum initial speed of the car with which the stunt driver must drive off the horizontal ramp to jump over eight cars parked below the ramp is $40{\mathrm{ms}}^{-1}$.

(b) The new minimum speed of the car with which it takes off from the ramp inclined at an angle $\theta =7.0°$is $23.9{\mathrm{ms}}^{-1}$.

Step-by-step solution

Step 1. Kinematic equations for projectile motion

If xandy are the horizontal and vertical displacements of the projectile traveled in time t, $v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$ are the horizontal and vertical components of the initial velocity of the particle, and $v_{\mathrm{x}},v_{\mathrm{y}}$ are the horizontal and vertical components of velocity of the particle at time t, then the kinematic equations for projectile motioncan be written as:

• For the horizontal motion of projectile:

$\begin{array}{c}{v}_{\mathrm{x}}=v_{\mathrm{x}0}\\ x=x_0+v_{\mathrm{x}0}t\end{array}$

• For the vertical motion of projectile:

$\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}+\mathrm{g}t\\ y=y_0+v_{\mathrm{y}0}t+\frac{1}{2}\mathrm{g}{t}^2\\ v_{\mathrm{y}}^2=v_{\mathrm{y}0}^2+2\mathrm{g}y\end{array}$

Here, the downward direction is taken as the positive direction. Therefore the vertical acceleration of the particle is taken to be equal to the acceleration due to gravity, i.e., $a=g$.

When the upward direction is taken as positive, g is replaced by –g in the above equations.

Step 2. Given information for part (a)

The minimum horizontal distance that must be covered by the car is $x=22\mathrm{m}$.

The vertical distance covered by the car is $y=1.5\mathrm{m}$.

If the downward direction is taken as the positive direction, then the vertical acceleration of the car will be equal to the acceleration due to gravity, i.e.,$a=\mathrm{g}$

Let the minimum initial speed of the car to drive off the horizontal ramp be $v_0$.

Step 3. (a) Determination of the minimum initial speed of the car

Here, your projectile is the car. Let t be the time taken by the car to jump over eight cars parked below the horizontal ramp and$v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$be the horizontal and vertical components of the initial velocity ${\overrightarrow{v}}_0$ of the car.

Since the car takes off along the horizontal ramp, the vertical component of the initial velocity will be zero, i.e., $v_{\mathrm{y}0}=0{\mathrm{ms}}^{-1}$. Thus, the initial speed of the car is $v_0=v_{\mathrm{x}0}$.

Using the kinematic equation for the vertical motion of the car, you will get time t as:

$\begin{array}{c}y=v_{y0}t+\frac{1}{2}g{t}^2\\ 1.5\mathrm{m}=\left(0{\mathrm{ms}}^{-1}\right)t+\frac{1}{2}\left(9.8{\mathrm{ms}}^{-2}\right)t^2\\ t^2=\frac{2\times \left(1.5\mathrm{m}\right)}{\left(9.8{\mathrm{ms}}^{-2}\right)}\\ t=0.306{\mathrm{s}}^2\\ t=0.55\mathrm{s}\end{array}$

Using the kinematic equation for the horizontal motion of the car, you will get:

$\begin{array}{c}x=v_{\mathrm{x}0}t\\ v_{\mathrm{x}0}=\frac{x}{t}\\ =\frac{\left(22\mathrm{m}\right)}{\left(0.55\mathrm{s}\right)}\\ =40{\mathrm{ms}}^{-1}\end{array}$

So, the minimum initial speed of the car with which the stunt driver must drive off the horizontal ramp to jump over eight cars parked below the ramp is$\mathbf{40}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$.

Step 4. Given information for part (b)

If the ramp is tilted upward at an angle $\theta =7.0°$then the initial velocity of the car will have non-zero horizontal and vertical components.

The horizontal and vertical components of the initial velocity ${{\overrightarrow{v}}_0}^{\text{'}}$of the car are:

$v_{\mathrm{x}0}={v_0}^{\text{'}}\cos 7.0°$

$v_{\mathrm{y}0}={v_0}^{\text{'}}\sin 7.0°$

If the upward direction is taken to be positive, then$y=-1.5\mathrm{m};a=-\mathrm{g}$

Let the time taken by the car to jump over the eight parked cars from the ramp inclined at an angle $\theta =7.0°$be$t\text{'}.$

Step 5. (b) Determination of the new minimum initial speed of the car when the ramp is tilted upward

Using the kinematic equation for vertical motion of the car, you will get:

$\begin{array}{c}y={v_{\mathrm{y}0}}^{\text{'}}t-\frac{1}{2}\mathrm{g}{t\text{'}}^2\\ -1.5=\left({v_0}^{\text{'}}\sin 7.0°\right)t\text{'}-\frac{1}{2}\left(9.8\right){t\text{'}}^2\dots \left(i\right)\end{array}$

Using the kinematic equation for the horizontal motion of the car, you will get:

$\begin{array}{c}x={v_{x0}}^{\text{'}}t\\ 22=\left({v_0}^{\text{'}}\cos 7.0°\right)t\text{'}\\ v_0=\frac{22}{\left(\cos 7.0°\right)t\text{'}}...\left(ii\right)\end{array}$

Substitute the value of $v_0$from (ii) in equation (i) to get the value of time $t\text{'}$.

$\begin{array}{c}-1.5=\left(\frac{22}{\left(\cos 7.0°\right)t}\sin 7.0°\right)t\text{'}-\frac{1}{2}\left(9.8\right){t\text{'}}^2\\ -1.5=22\tan 7.0°-4.9{t\text{'}}^2\\ 4.9{t\text{'}}^2=22\tan 7.0°+1.5\\ {t\text{'}}^2=0.86\\ t\text{'}=0.93\mathrm{s}\end{array}$

Thus, thetime taken by the car to jump off eight parked cars from the ramp inclined at an angle$\theta =\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{°}$is$\mathbf{0}\mathbf{.}\mathbf{93}\mathbf{s}$.

Substitute the value of t in equation (ii) to get the value of $v_0$.

$\begin{array}{c}{v_0}^{\text{'}}=\frac{22}{\left(\cos 7.0°\right)\left(0.93\mathrm{s}\right)}\\ =\frac{22}{\left(0.99\right)\left(0.93\mathrm{s}\right)}\\ =23.9{\mathrm{ms}}^{-1}\end{array}$

Thus, the new minimum speed of the car with which it takes off from the ramp inclined at an angle$\theta \mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{°}$is$\mathbf{23}\mathbf{.}\mathbf{9}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$.