Car Stunt: Driving off a Cliff
How can we calculate the distance a car travels after being driven off a cliff?
A stunt man drives a car at a speed of 20 m/s off a 25-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20∘. How far from the base of the cliff does the car land?
Answer:
The problem entails using kinematics to solve for the distance a car travels after being driven off a cliff at a certain angle. It calls for the calculation of the time the car spends in the air and the horizontal distance it travels during that time. The answer is approximately 41.63 meters.
To answer the student's question, we'll utilize principles of kinematics which involves the study of motion in physics. Since the car leaves the cliff at an angle of 20 degrees, both horizontal and vertical motions need to be considered. The vertical height of the cliff, that is 25 meters, can be used to calculate the time the car spends in the air using the formula: 1/2 * g * t^2 = height, where g is the acceleration due to gravity (9.8 m/s^2). Solving for t, we get about 2.27 seconds. The horizontal distance the car travels can be achieved by calculating the horizontal component of the car's initial velocity and multiplying it by the time spent in the air. That is, distance = (initial velocity * cos(20)) * time = (20 m/s * cos(20)) * 2.27s = approx. 41.63 meters. Therefore, the car would land approximately 41.63 meters away from the cliff's base.