Q63 P - University Physics with Modern Physics

Question

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. P3.63). The take off ramp was inclined at $\mathbf{53}\mathbf{°}$, the river was $\mathbf{40}\mathbf{.}\mathbf{0}\mathit{m}$ wide, and the far bank was $\mathbf{15}\mathbf{.}\mathbf{0}\mathit{m}$  lower than the top of the ramp. The river itself was $\mathbf{100}\mathit{m}$ below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

Step-by-Step Solution

Answer

(a) Speed  $u=17.8m/s$

 

(b) The landing distance $s=28.4\text{m}$

Step 1: Newton’s law of motion

The velocity contains two components, one is a horizontal component and another one is vertical.

 

According to newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, s, u, t, a, are displacement, initial velocity, time, and acceleration.

For the vertical and horizontal motion, the velocity component will be $u\sin \theta \text{ and cos}\theta$ respectively.

Step 2: Given

Angle =  $53°$

 

The wideness of the river =  $\text{40m}$

 

Far bank deepness =   $\text{15m}$

                                                                                                                        

The river below the ramp = $\text{100m}$

Step 3: (a) Speed at the top of the ramp

From the horizontal motion, gravity for horizontal motion will be zero.

$$\begin{gathered} s=u\cos \theta t+\frac{1}{2}a{t}^2 \\ 40\text{m}=u\cos 53°t \\ ut=66.47\text{m} \end{gathered}$$

From the vertical motion,

$$\begin{gathered} s=u\sin \theta t+\frac{1}{2}a{t}^2 \\ \text{ - 15.0m}=66.47\text{m}\times \text{sin}53°+\frac{1}{2}\times -9.8m/s^2\times t^2 \\ t=3.727\text{s} \end{gathered}$$

Now the speed will be,

$$\begin{gathered} ut=66.47\text{m} \\ u=\frac{66.47\text{m}}{3.727\text{s}} \\ u=17.8 m/s \end{gathered}$$

Step 4: (b) Landing distance with half of the speed

$$\begin{gathered} u=\frac{17.8m/s}{2} \\ u=8.9 m/s \end{gathered}$$

From the vertical motion,

$$\begin{gathered} s=u\sin \theta t+\frac{1}{2}a{t}^2 \\ \text{ - 100.0m}=8.9 m/s \times \text{sin}53°t+\frac{1}{2}\times -9.8 m/s^2\times t^2 \\ t=5.30\text{s} \end{gathered}$$

From the horizontal motion, gravity for horizontal motion will be zero.

$$\begin{gathered} s=u\cos \theta t+\frac{1}{2}a{t}^2 \\ s=8.9 m/s\times \cos 53°\times 5.30\text{s} \\ s=28.4\text{m} \end{gathered}$$

The  Speed at the top of the ramp will be $u=17.8 m/s$ and the landing distance will be  $s=28.4\text{m}$.