a car is going 8 m/s goes over a cliff 78.4 meters high. How far from the base does it land?
Calculate how long it takes to fall 78.4 m.
78.4 m = (g/2) t^2
Solve for t.
Multiply that t by 8 m/s for the answer.,
32
To answer this question, we can use the equations of motion to determine the distance from the base where the car lands. The equation that relates distance, initial velocity, time, and acceleration is:
d = vt + (1/2)at^2
Where:
- d represents the distance traveled
- v is the initial velocity (8 m/s in this case)
- t is the time
- a is the acceleration (in this case, the acceleration due to gravity, approximately 9.8 m/s^2)
First, let's calculate the time it takes for the car to reach the ground. We know that the initial velocity is 8 m/s and the distance traveled vertically is the height of the cliff, 78.4 meters. The equation we can use to find the time is:
d = vt + (1/2)at^2
Rearranging the equation and substituting the given values, we have:
78.4 = (8)t + (1/2)(9.8)t^2
Now we can solve this quadratic equation for t.
(1/2)(9.8)t^2 + 8t - 78.4 = 0
Solving this equation, we find two possible solutions: t ≈ 4.2 seconds and t ≈ -8.1 seconds. Since time cannot be negative in this context, we disregard the negative solution, and the car takes approximately 4.2 seconds to reach the ground.
Now that we know the time it takes, we can find the horizontal distance the car travels using the equation:
d = vt
Substituting the values for v and t:
d = 8(4.2) = 33.6 meters
Therefore, the car lands approximately 33.6 meters from the base of the cliff.