Question

A ball is thrown horizontally off a cliff with an initial velocity of 20 m/s. The cliff is 100 meters high. How far from the base of the cliff will the ball land? Assume air resistance is negligible.

Answer 1

To solve this problem, we can use the equations of motion for a horizontally launched projectile. Since the ball is thrown horizontally, the initial vertical velocity is zero. The key equation we'll use is:

$s=ut+\frac{1}{2}a{t}^2$

where: s = vertical displacement u = initial vertical velocity (0 m/s) a = acceleration due to gravity (-9.8 m/s^2, assuming downward as the positive direction) t = time

We can calculate the time it takes for the ball to reach the ground using the equation:

$s=ut+\frac{1}{2}a{t}^2$

$0=(0\text{m/s})\cdot t+\frac{1}{2}\cdot (-9.8{\text{m/s}}^2)\cdot t^2$

Simplifying the equation:

$0=-4.9t^2$

Since the time cannot be negative, we can discard the trivial solution t = 0. Thus, we have:

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