A rock is thrown from the edge of a cliff to the ground 19.5 m below. The rock h

Question

A rock is thrown from the edge of a cliff to the ground 19.5 m below. The rock has an initial velocity of 17.5 m/s, directed 16.5° above the horizontal.
How long does it take the rock to reach the ground?
How far from the base of the cliff does the rock strike the ground?
Find the velocity of the rock just before it strikes the ground: magnitude=
direction= (below the horizontal)


I know this question has a lot to it and if you can't answer it all that's okay. I just have been working on it for an hour now and have a test tomorrow morning on this subject. I would really appreciate any help I can get!

Explanation / Answer

To get the time it takes for the rock to hit the ground, you need the initial y component of the rock's velocity: sin(angle) = Vy / V sin(16.5 degrees) = Vy / 17.5 Vy = 4.97 m/s Now, you can find the time by using the following equation (since you know the acceleration, initial speed, and total distance) d = Vi*t + (1/2)*a*t^2 -19.5 = 4.97 * t + (1/2) * (-9.8) * t^2 using the quadratic equation, you can solve for t and get t = 2.57 seconds. So it takes the rock 2.57 seconds to reach the ground. To find the distance from the base of the cliff, you need the x component of the initial rock velocity: cos(angle) = Vx / V cos(16.5 degrees) = Vx / 17.5 Vx = 16.78 m/s Now, knowing that the x component velocity does not change through the flight of the rock, you can just multiply the time by the initial x velocity to get the horizontal distance traveled: D = Vx * t, D = 16.78 * 2.57, D = 43.12 meters from the base of the cliff Lastly, to get the velocity you just need the final y component of the velocity (since you already have the final x component since it doesn't change) Vf = Vi + a*t Vf = 4.97 + (-9.8) * 2.57 Vf = -20.22 m/s To get the magnitude of the total final velocity, you can use: Mag^2 = Vy^2 + Vx^2 = (-20.22)^2 + 16.78^2, so Mag = 26.28 m/s To get the direction, you can use SOHCAHTOA: tan(angle) = Vy / Vx tan(angle) = 20.22 / 16.78 angle = 50.31 degrees So the direction is 50.31 degrees below the horizontal.

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