A car drives straight off the edge of a cliff that is 58 m high. The police at the scene of the accident note that the point of impact is 126 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
To determine the speed of the car when it went over the cliff, we can use the principles of projectile motion.
Let's break down the information given in the problem:
- Height of the cliff (h) = 58 m
- Horizontal distance from the base of the cliff to the point of impact (d) = 126 m
To solve this problem, we need to find the initial velocity (speed) of the car when it went off the cliff, also known as the horizontal component of the velocity (v₀x).
Here's the step-by-step process to find the speed of the car:
Step 1: Identify the relevant equation
The key equation that relates distance, time, and initial velocity in projectile motion is:
d = v₀x * t (Equation 1)
where:
- d is the horizontal distance traveled (126 m in this case),
- v₀x is the initial horizontal velocity of the car, and
- t is the time the car spends in the air.
Step 2: Determine time in the air
Since the car falls vertically, the time it spends in the air can be found using the vertical component of the motion. The equation for the vertical displacement in the absence of air resistance is:
h = (1/2) * g * t² (Equation 2),
where:
- h is the height of the cliff (58 m in this case),
- g is the acceleration due to gravity (approximately 9.8 m/s²), and
- t is the time of flight.
Using Equation 2, we can solve for time (t):
2h = g * t²
t² = (2h) / g
t = √[(2h) / g]
Step 3: Calculate the time of flight
Substitute the given values into the equation we derived for time (t):
t = √[(2 * 58) / 9.8]
t ≈ √[11.84]
t ≈ 3.44 s (rounded to two decimal places)
Step 4: Calculate the initial horizontal velocity
Now, we can substitute the obtained time (t) into Equation 1:
d = v₀x * t
v₀x = d / t
v₀x = 126 m / 3.44 s
Calculating v₀x:
v₀x ≈ 36.63 m/s (rounded to two decimal places)
Thus, the car was traveling at a speed of approximately 36.63 m/s when it went over the cliff.