Question

A stone is thrown at an angle of 30° above the horizontal from the edge of a cliff with an initial speed of 12 m/s. A stopwatch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff?

Answer 1

To solve this problem, we can use the kinematic equations of motion.

1. The vertical component of the initial velocity can be calculated using sin(30°) = vertical velocity / initial velocity. Therefore, the vertical component of the initial velocity is 12 m/s * sin(30°) = 6 m/s.

2. We know that the acceleration due to gravity is -9.8 m/s^2 for the upward direction. Using the equation, final height = initial height + initial vertical velocity * time + 0.5 * acceleration * time^2, we can plug in the known values to solve for the height:

final height = 0 + 6 m/s * 5.6 s + 0.5 * -9.8 m/s^2 * (5.6 s)^2
final height = 33.6 m + 0.5 * -9.8 m/s^2 * 31.36 s^2
final height = 33.6 m - 153.808 m
final height = -120.208 m

Since the height cannot be negative, it means that the initial guess for the initial vertical velocity was incorrect. The correct vertical component of the initial velocity should be -6 m/s. Using this value:

final height = 0 + (-6 m/s) * 5.6 s + 0.5 * -9.8 m/s^2 * (5.6 s)^2
final height = -33.6 m + 0.5 * -9.8 m/s^2 * 31.36 s^2
final height = -33.6 m + (-153.808 m)
final height = -187.408 m

Therefore, the height of the cliff is approximately 187.4 meters.