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.

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

Given the following data:

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:

Making t the subject of formula, we have:

Substituting the given parameters into the formula, we have;

Time, t = 1.99 seconds

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

Horizontal speed = 13.22 m/s

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The launch speed is approximately 13.28 m/s

Explanation:

The given parameters are;

The distance from the base of the cliff the projectile lands = 26.3 meters

The height of the cliff, s = 19.3 meters

The time it takes the projectile to land = 5 seconds

The direction of the projectile = Horizontal

The vertical motion of the projectile is given as follows;

19.3 = 1/2 × g × t² = 1/2 × 9.81 × t²

t = √(2 × 19.3/9.81) ≈ 1.98 seconds

Therefore, the time it takes to land ≈ 1.98 seconds

The launch (horizontal) speed = (The horizontal distance)/(Horizontal time)

The launch (horizontal) speed = 26.3/1.98 ≈ 13.28 m/s.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

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