A stunt car driver drives off an
h = 6.7 m-high cliff into a lake with a horizontal speed of 14 m/s. The car needs to clear an 8.0 m-long ledge that is 5.0 m below the edge of the cliff. The car successfully misses the ledge on the way down. With what speed does the car hit the water?
initial KE+initial PE=final KE
1/2 m 14^2 + mg*6.7=1/2 m vf^2
vf^2=1/2 * 228+9.8*6.7
solve vf
To find the speed at which the car hits the water, we can use the principle of conservation of energy. At the top of the cliff, the car has gravitational potential energy given by mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the cliff.
At the bottom of the cliff, just before hitting the water, the car has both kinetic energy and gravitational potential energy. The kinetic energy is given by (1/2)mv^2, where v is the velocity of the car just before hitting the water.
Since energy is conserved, we can equate the initial potential energy at the top of the cliff to the final combination of kinetic energy and potential energy just before hitting the water.
mgh = (1/2)mv^2 + mgh'
Where h' is the height from the water surface to the bottom of the cliff.
To solve for v, we can cancel out the mass on both sides of the equation:
gh = (1/2)v^2 + gh'
Now let's plug in the given values:
g = 9.8 m/s^2
h = 6.7 m
h' = 5.0 m
gh = (1/2)v^2 + gh'
(9.8 m/s^2)(6.7 m) = (1/2)v^2 + (9.8 m/s^2)(5.0 m)
Now, for simplicity, let's solve for v^2:
v^2 = 2(9.8 m/s^2)(6.7 m + 5.0 m) - which is 2 * (9.8 m/s^2) (11.7m)
v^2 = 2 * (9.8 m/s^2) * (11.7m)
v^2 = 22.47 m^2/s^2
Finally, taking the square root of both sides to solve for v:
v = sqrt(22.47 m^2/s^2)
Therefore, the speed at which the car hits the water is approximately 4.74 m/s.