Question

a car comes to a bridge during a storm and finds the bridge washed out. the driver must get to the other side, so he decides to try leaping it with his car. the side the car is on is 18.8 m above the river, whereas the opposite side is a mere 1.6 m above the river. the river itself is a raging torrent 50.0 m wide. for help with math skills, you may want to review vector magnitudes. for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of different initial and final heights.
part b
what is the speed of the car just before it lands safely on the other side?
express your answer in meters per second.
view available hint(s)
( v = ) (\boxed{}) ( \frac{m}{s} )

Explanation:

Step1: Analyze vertical motion (free fall)

The vertical displacement is \( \Delta y = 18.8 - 1.6 = 17.2 \, \text{m} \) (downward, so positive). Use the equation \( \Delta y = v_{0y}t+\frac{1}{2}gt^2 \). Initial vertical velocity \( v_{0y} = 0 \, \text{m/s} \) (horizontal leap), so \( \Delta y=\frac{1}{2}gt^2 \). Solve for time \( t \):
\( t = \sqrt{\frac{2\Delta y}{g}} \), where \( g = 9.8 \, \text{m/s}^2 \).
\( t=\sqrt{\frac{2\times17.2}{9.8}}=\sqrt{\frac{34.4}{9.8}}\approx\sqrt{3.51}\approx1.87 \, \text{s} \).

Step2: Analyze horizontal motion (constant velocity)

Horizontal displacement \( \Delta x = 50.0 \, \text{m} \), time \( t \approx 1.87 \, \text{s} \). Horizontal velocity \( v_x=\frac{\Delta x}{t} \) (constant, no air resistance).
\( v_x=\frac{50.0}{1.87}\approx26.7 \, \text{m/s} \).

Step3: Find final speed (magnitude of velocity vector)

Final vertical velocity \( v_y = v_{0y}+gt = 0 + 9.8\times1.87\approx18.3 \, \text{m/s} \).
Speed \( v=\sqrt{v_x^2 + v_y^2}=\sqrt{(26.7)^2+(18.3)^2}=\sqrt{712.89 + 334.89}=\sqrt{1047.78}\approx32.4 \, \text{m/s} \).

Answer:

\( \approx 32.4 \, \text{m/s} \)