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. how fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side? express your answer in meters per second. view available hint(s) v₀ = m/s

Explanation:

Step1: Analyze vertical motion (free - fall)

The vertical displacement of the car is the difference in height between the two sides of the river. The initial height \(y_0 = 18.8\space m\) and the final height \(y=1.6\space m\), so the vertical displacement \(\Delta y=y - y_0=1.6 - 18.8=- 17.2\space m\) (negative because it is downward). The initial vertical velocity \(v_{0y} = 0\space m/s\) (since the car is moving horizontally when it leaves the bridge), and the acceleration due to gravity \(g = 9.8\space m/s^{2}\).

We use the equation of motion for vertical displacement: \(\Delta y=v_{0y}t+\frac{1}{2}at^{2}\). Substituting \(v_{0y} = 0\) and \(a=-g\) (taking upward as positive), we get \(\Delta y=-\frac{1}{2}gt^{2}\).

We can solve for time \(t\):
\[t=\sqrt{\frac{- 2\Delta y}{g}}\]
Substituting \(\Delta y=- 17.2\space m\) and \(g = 9.8\space m/s^{2}\):
\[t=\sqrt{\frac{-2\times(- 17.2)}{9.8}}=\sqrt{\frac{34.4}{9.8}}\approx\sqrt{3.51}\approx1.87\space s\]

Step2: Analyze horizontal motion (uniform motion)

In horizontal direction, there is no acceleration (\(a_x = 0\)), so the horizontal displacement \(x = v_{0x}t\), where \(x = 50.0\space m\) (width of the river) and \(t\) is the time we found in step 1. We need to find the initial horizontal velocity \(v_{0x}\) (which is the speed of the car when it leaves the bridge).

From \(x = v_{0x}t\), we can re - arrange to get \(v_{0x}=\frac{x}{t}\)

Substituting \(x = 50.0\space m\) and \(t\approx1.87\space s\):
\[v_{0x}=\frac{50.0}{1.87}\approx26.7\space m/s\]

Answer:

\(26.7\space m/s\) (approximate value, the more precise calculation of \(t=\sqrt{\frac{34.4}{9.8}}\approx1.873\space s\), then \(v_{0x}=\frac{50}{1.873}\approx26.7\space m/s\))