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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 4.0 s later.

User Riri
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Answer: I am assuming you're asking for the cliff height, which is 24M

We can use the kinematic equations to find the horizontal and vertical components of the ball's velocity and its final position.

The initial velocity components are:

v_x = 33 m/s * cos(60∘) = 28.6 m/s

v_y = 33 m/s * sin(60∘) = 33 * √3/2 m/s

The displacement components can be found using the kinematic equation:

x = v_x * t = 28.6 m/s * 4.0 s = 114.4 m

y = v_y * t - 0.5 * g * t^2 = 33 * √3/2 m/s * 4.0 s - 0.5 * 9.8 m/s^2 * (4.0 s)^2 = 102.4 m - 78.4 m = 24.0 m

The ball lands on the edge of the cliff, so its height must be equal to h, or:

y = h

24.0 m = h

So, the height of the cliff is 24.0 m.

User Andrea Zonca
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