Question

A 0.650 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1950 J and at its highest point is 131.8 m above the launch point.

(a) What is the horizontal component of its velocity?

(b) What was the vertical component of its velocity just after launch?

(c) At one instant during its flight the vertical component of its velocity is 25.41 m/s. At that time, how far is it above or below the launch point?

Answer 1

Let’s use the initial kinetic energy to determine the object’s initial velocity.
½ * 0.65 * v^2 =1950
v^2 = 1950/0.325
v = √(1950/0.325) ≈ 77.459 m/s

As the object moves up to its maximum height, its vertical velocity decreases to 0 m/s. Use the following equation to determine the object’s initial vertical velocity.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -9.8, d = 131.8
0 = vi^2 + 2 * -9.8 * 131.8
-vi^2 = -2583.28
vi = √2583.28 = 50.83 m/s
This is the initial vertical velocity.

Initial vertical velocity = v * sin θ
√(1950/0.325) * sin θ = √2583.28
sin θ = 0.6561
The angle is approximately 41˚

Horizontal component = v * cos θ = √(1950/0.325) * cos 41˚
The horizontal component is approximately 58.45 m/s

c) At the instant the vertical component of its velocity is 25.41 m/s, what is its vertical displacement from the launch point? Take up to be the positive direction

Use the following equation to determine the vertical displacement from the launch point.


vf^2 = vi^2 + 2 * a * d
vf = 25.41, vi = 50.83, a = -9.8
25.41^2 = 50.83^2 + 2 * -9.8 * d

-1938.02 = -19.6 * d
d = 98.87 m

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