In a movie scene involving a car chase, a car goes over the top of a ramp at A and lands at B below. Letting ˛α=18° and β=25°, determine the speed of the car at A if the car is to be airborne for a full 3 s. Furthermore, determine the distance d covered by the car during the stunt as well as the impact speed and angle at B. Neglect aerodynamic effects. Express your answer using the U.S. Customary system of units.

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Accepted Answer

This problem is a projectile problem and we begin by relating the jump speed [latex]v_0[/latex] to the velocity of the car and then to the car’s trajectory. Referring tot he figure on the right, the initial velocity and acceleration in the (x , y) coordinate system are

[latex]\begin{aligned}\dot{x}_0 & =v_0 \cos (\alpha+\beta), & \dot{y}_0 & =v_0 \sin (\alpha+\beta), \\ddot{x} & =g \sin \beta, & \ddot{y} & =-g \cos \beta .\end{aligned}[/latex]

Using the constant acceleration equation [latex]s=\dot{s} t+\frac{1}{2} a_c t^2[/latex] , we have that the coordinates of the (airborne) car as a function of time are

[latex]\begin{aligned}& x=v_0 \cos (\alpha+\beta) t+\frac{1}{2} g \sin \ \beta \ t^2, \quad ext{(1)}\& y=v_0 \sin (\alpha+\beta) t-\frac{1}{2} g \cos \ \beta \ t^2. \quad ext{(2)}\end{aligned}[/latex]

We want to satisfy the condition that y = 0 at t = [latex]t_f[/latex] = 3 s, where the subscript f stands for flight.

[latex]0=v_0 \sin (\alpha+\beta) t_f-\frac{1}{2} g \cos \ \beta \ t_f^2 \quad \Rightarrow \quad v_0=\frac{g t_f \cos \beta}{2 \sin (\alpha+\beta)}=64.19 \ ft / s.[/latex] (3)

where we have used the following numerical data: g = 32.2 ft/s² ,[ latex]t_f[/latex]= 3 s, β = 25° , and α = 18°. Expressing the result for [latex]v_0[/latex] to 3 significant figures, we have

[latex]v_0=64.2 \ ft / s.[/latex]

We now observe that the distance d is equal to the value of x at time t = [latex]t_f[/latex] . Hence, using Eq. (1) and the value of [latex]v_0[/latex] we computed in the last of Eqs. (3), we have that the distance d is given by

[latex]d=\frac{g t_f \cos \beta}{2 \sin (\alpha+\beta)} \cos (\alpha+\beta) t_f+\frac{1}{2} g \sin \ \beta \ t_f^2=202 \ ft .[/latex]

To find the impact speed and the impact angle, we need to determine the velocity at impact. To do so, we use the constant acceleration equation [latex]v=v_0+a_c t[/latex] act and we obtain

[latex]\vec{v}_i=\left[v_0 \cos (\alpha+\beta)+g \sin \ \beta \ t_f ight] \hat{\imath}+\left[v_0 \sin (\alpha+\beta)-g \cos \ \beta \ t_f ight] \hat{\jmath}=(87.77 \hat{\imath}-43.77 \hat{\jmath}) ft / s[/latex] (4)

where we have used the value of [latex]v_0[/latex] in the last of Eqs. (3) along with the following numerical data: g = 32.2 ft / s² , [latex]t_f[/latex] = 3 s, β = 25° , and α = 18° . Therefore the speed at impact is given by

[latex]v_i=\sqrt{v_{x i}^2+v_{y i}^2}=98.1 \ ft / s.[/latex]

We now let [latex]\phi[/latex] be the impact angle of the car with the ground at B, i.e., with respect to the slope. Hence, this angle is given by

[latex]\phi= an ^{-1} \frac{-v_y}{v_x}=26.5^{\circ} ext { measured clockwise from the slope. }[/latex]

Question 2.241: In a movie scene involving a car chase, a car goes over the ...

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