WichoSolis04
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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.​

Respuesta :

Answer:

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.

Lanuel

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 meters
  • Horizontal 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:

[tex]H = \frac{1}{2} gt^2[/tex]

Making t the subject of formula, we have:

[tex]t=\sqrt{\frac{2H}{g} }[/tex]

Substituting the given parameters into the formula, we have;

[tex]t=\sqrt{\frac{2\times 19.3}{9.8} }\\\\t=\sqrt{\frac{38.6}{9.8} }\\\\t=\sqrt{3.94}[/tex]

Time, t = 1.99 seconds

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

[tex]Horizontal\;speed = \frac{horizontal\;distance}{time} \\\\Horizontal\;speed = \frac{26.3}{1.99}[/tex]

Horizontal speed = 13.22 m/s

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