A car drives straight off the edge of a cliff that is \(60.0 \mathrm{~m}\) high. The police at the scene of the accident note that the point of impact is \(150 . \mathrm{m}\) from the base of the cliff. How fast was the car traveling when it went over the cliff?

Short Answer

Expert verified
Answer: The car was traveling at approximately 42.87 m/s when it went over the cliff.

Step by step solution

01

Determine the time

Since the vertical motion is uniformly accelerated, we can use the following equation to find the time it takes for the car to fall to the ground: \(h = \frac{1}{2}gt^2\) where \(h\) is the height of the cliff, \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)), and \(t\) is the time it takes for the car to fall. Solving for time: \(t = \sqrt{\frac{2h}{g}}\) Substituting the given values: \(t = \sqrt{\frac{2(60)}{9.81}}\)
02

Calculate time using the given values

Calculate the time based on the height of the cliff and the acceleration due to gravity: \(t = \sqrt{\frac{120}{9.81}} \approx 3.499 \mathrm{~s}\)
03

Determine the car's initial horizontal speed

Since the horizontal motion is uniform, we can use the following equation to find the initial horizontal speed of the car: \(x = vt\) where \(x\) is the horizontal distance and \(v\) is the initial horizontal speed of the car. Solving for the initial horizontal speed: \(v = \frac{x}{t}\) Substituting the given values: \(v = \frac{150}{3.499}\)
04

Calculate the speed of the car

Calculate the initial speed of the car using the horizontal distance and the time it took for the car to hit the ground: \(v = \frac{150}{3.499} \approx 42.87 \mathrm{~m/s}\) The car was traveling at approximately \(42.87 \mathrm{~m/s}\) when it went over the cliff.

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