Question

A ball is thrown horizontally off the edge of a cliff with an initial speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How long does it take for the ball to fall to the ground 20.0 m below? (b) How long would it take for the ball to reach the ground if it were dropped from rest off the cliff edge? (c) How long would it take the ball to fall to the ground if it were thrown at an initial velocity of \(20.0 \mathrm{m} / \mathrm{s}\) but \(18^{\circ}\) below the horizontal?

Short answer

Rearranging the equation and solving for \(t\) gives: \(t^2 = \dfrac{2(20)}{9.81}\) \(t^2 ≈ 4.08\) Therefore, \(t = \sqrt{4.08} ≈ 2.02 \mathrm{s}\) In scenario (a) where the ball is thrown horizontally, it takes approximately 2.02 seconds for the ball to hit the ground.

Step-by-step solution

Scenario (a): Ball thrown horizontally

In this scenario, the ball is thrown horizontally off the edge of the cliff with an initial speed of 20.0 m/s. Since the motion of the ball is happening both horizontally and vertically, we'll have separate equations for horizontal and vertical planes. First, we need to find the time it takes for the ball to fall 20.0 m below. Using the vertical motion equation with an initial velocity of 0 (due to no vertical motion initially) and with acceleration due to gravity \(\mathrm{g} = 9.81 \mathrm{m} / \mathrm{s}^2\), we have: \(y = v_{0y} t - 0.5 gt^2\) Plugging in the values, we get: \(20 = 0 - 0.5(9.81)t^2\) Now, we need to solve for \(t\).