Question

A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160m from the base of the building 5.0s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with whichthe rock was thrown.

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration \(=\mathrm{g}=-9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Initial speed of the rock \(=\mathrm{V}\)

Initial horizontal speed of the rock \(=V_{x}\)

Initial vertical speed of the rock \(=V_{y}\)

Height of the building \(=\mathrm{H}=100 \mathrm{~m}\)

Horizontal distance covered by the rock before hitting the ground \(=\mathrm{R}=160 \mathrm{~m}\)

Time taken by the rock to hit the ground \(=\mathrm{T}=5 \mathrm{sec}\)

There is no horizontal force on the rock therefore the horizontal velocity of the rock remains constant.

$$ \begin{aligned} &\mathrm{R}=\mathrm{V}_{\mathrm{x}} \mathrm{T} \\ &160=\mathrm{V}_{\mathrm{x}}(5) \\ &\mathrm{V}_{\mathrm{x}}=32 \mathrm{~m} / \mathrm{s} \end{aligned} $$

The displacement of the rock in the vertical direction when it hits the ground is in the downward direction therefore it is negative,

$$ \begin{aligned} &-\mathrm{H}=V_{\mathrm{y}} \mathrm{T}+\mathrm{gT}^{2} / 2 \\ &-100=\mathrm{V}_{y}(5)+(-9.81)(5)^{2} / 2 \\ &\mathrm{~V}_{\mathrm{y}}=4.525 \mathrm{~m} / \mathrm{s} \\ &V=\sqrt{\left(V_{x}\right)^{2}+\left(V_{y}\right)^{2}} \\ &V=\sqrt{(32)^{2}+(4.525)^{2}} \\ &\mathrm{~V}=32.32 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Speed with which the rock was thrown \(=32.32 \mathrm{~m} / \mathrm{s}\)