A student stands at the edge of a cliff and throws a stone horizontally over the
ID: 1682841 • Letter: A
Question
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 23.0 m/s. The cliff is h = 24.0 m above a flat horizontal beach, as shown in the figure below.How long after being released does the stone strike the beach below the cliff?
1 s
With what speed and angle of impact does the stone land?
2 m/s
3° below the horizontal
Explanation / Answer
The time taken by the body to reach the ground is h = 0.5 gt^2 ==> t = [2 h / g]^1/2 = [2 * 24 / 9.8]^1/2 = 2.213 s The final speed of the stone v = [Vx^2 + Vy^2]^1/2 = [Vx^2+g^2*t^2]^1/2 = [23^2 + (9.8 * 2.213)^2]1/2 = 31.61 m/s angle theta = tan-1 ( Vy/Vx) = tan-1 (9.8 * 2.212 / 23) = 43.31 deg
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.