A stunt car driver drives off an
h = 7.9 m-high cliff into a lake with a horizontal speed of 17 m/s. The car needs to clear an 17.1 m-long ledge that is 5.0 m below the edge of the cliff, as shown in the figure below. The car just misses the ledge on the way down. With what speed does the car hit the water?
I do not see the figure but am at a loss to understand how the ledge has anything to do with it unless he hits it.
u = 17 m/s until splash
Vi = 0
h = 7.9 m
h = 4.9 t^2
so
t = 1.27 seconds
v = 9.81 t = 9.81*1.27 = 12.5 m/s
speed = sqrt (17^2 + 12.5^2)
To find the speed at which the car hits the water, we need to analyze the vertical and horizontal components of its motion separately.
First, let's find the time it takes for the car to fall from the cliff to the lake. We can use the equation for the vertical displacement:
h = (1/2) * g * t^2
Where:
h = vertical displacement (7.9 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation to solve for t, we get:
t = sqrt((2 * h) / g)
Substituting the given values, we calculate:
t = sqrt((2 * 7.9) / 9.8) ≈ 1.41 seconds
Now, we need to find the horizontal distance the car travels during this time. We can use the equation for horizontal distance:
d = v * t
Where:
d = horizontal distance (17.1 m)
v = horizontal velocity (17 m/s)
t = time (1.41 s)
Rearranging the equation to solve for v, we get:
v = d / t
Substituting the given values, we calculate:
v = 17.1 / 1.41 ≈ 12.13 m/s
Therefore, the speed at which the car hits the water is approximately 12.13 m/s.