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?
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.