A student stands at the edge of a cliff and throws a stone horizontally over the

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 21.0 m/s. The cliff is h = 58.0 m above a flat, horizontal beach as shown in the figure.

vx = vy = A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 21.0 m/s. The cliff is h = 58.0 m above a flat, horizontal beach as shown in the figure. Write the equations for the x- and y-components of the velocity of the stone with time. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.) Write the equations for the position of the stone with time, using the coordinates in the figure. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.) How long after being released does the stone strike the beach below the cliff? With what speed and angle of impact does the stone land?

Explanation / Answer

In this case (or almost all projectile motion problems), you can decompose the motion of the ball into vertical and horizontal motion. They can be considered as completely separate.

A) Using the formula: s = (1/2)at^2
rearrange: t = squareroot(2s/a) = squareroot (2*58/9.8) = 3.44 seconds

B) The 21m/s in the horizontal direction remains constant.

Calculate the vertical speed when it hits the bottom.
v^2 = u^2 + 2as
v = squareroot(u^2 + 2as) = squareroot (0 + 2*9.8*58) = 33.7m/s

The 2 velocities will have to be added vectorially.

The speed is = squareroot(33.7^2 + 21^2) = 39.7m/s

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