Physics, 22.10.2020 19:01 theresaeleanor

Determine the launch speed of a horizontally launched projectile that lands 26.3 meters from the base of a 19.3-meter high cliff and takes 5 seconds to land.

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Answer from: xandraeden32

The launch speed of a horizontally launched projectile is equal to 13.22 m/s.

Given the following data:

Time = 5 seconds.Height = 19.3 metersHorizontal distance = 26.3 meters.

To determine the launch speed of a horizontally launched projectile:

First of all, we would calculate the time it took the projectile to land at the base of the cliff by using the maximum height formula:

$H = \frac{1}{2} gt^2$

Making t the subject of formula, we have:

$t=\sqrt{\frac{2H}{g} }$

Substituting the given parameters into the formula, we have;

$t=\sqrt{\frac{2\times 19.3}{9.8} }\\\\t=\sqrt{\frac{38.6}{9.8} }\\\\t=\sqrt{3.94}$

Time, t = 1.99 seconds

Next, we would determine the launch speed by using the formula:

$Horizontal\;speed = \frac{horizontal\;distance}{time} \\\\Horizontal\;speed = \frac{26.3}{1.99}$

Horizontal speed = 13.22 m/s

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