A rock is tossed straight up with a velocity of 14 m/s When it returns, it falls

Question

A rock is tossed straight up with a velocity of 14 m/s When it returns, it falls into a hole 10m deep.

What is the rock's velocity as it hits the bottom of the hole?

How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Explanation / Answer

This question can be easily solved using the 3 basic kinematic equations. These are: v = u + at s = vt + 1/2at^2 v^2 = u^2 + 2as Where: v = final velocity u = initial velocity s = displacement a = acceleration t = time These equations can give us the answer your looking for. Q. What is the rock's velocity as it hits the bottom of the hole? To answer this we first need to know what height it is falling from. This first thing to do would be to draw a diagram as this helps immensely when trying to understand the question. Now analyze what what info you are given: You know the rock leaves the ground @ 20m/s You know that when the rock reaches it's peak its velocity is 0 m/s as it neither moving up or down You know that gravity is working against the rock @ 9.8 m/s/s (thats acceleration) down (air resistance is disregarded) So you have: v = 20 m/s u = 0 m/s a = -9.8 m/s/s (assuming that upwards is a positive direction) and we want to find displacement (s), from this we can use the v^2 = u^2 + 2as equation as it takes into account all the factors we have as well as giving the displacement, which is what we want to find. So first thing is to rearrange the equation: v^2 = u^2 + 2as v^2 - u^2 = 2as (v^2 - u^2) / 2a = s So now input parameters: s = (20^2 - 0^2) / (2 * -9.8) s = 400 / -19.6 s = -20.408 As displacement does not have a direction, take the absolute value (take the negative away) s = 20.408 meters So the rock reaches its peak height at 20 meters off the ground, however it falls into a 10m deep hole below the ground from this height so then the displacement of the rock from its peak point (20.4 meters off the ground) would be 30.4 meters. We now know: s = 30.4m u = 0 m/s a = 9.8 m/s/s and we want to find velocity, v. Looking at the set of kinematic equations the last equation once again looks suitable for the job, as we have displacement, initial velocity and the acceleration but need to find the final velocity. Seeing as the equation is already in terms of final velocity, no rearranging is necessary. v^2 = u^2 + 2as v^2 = 0^2 + 2 * 9.8 * 30.4 v^2 = 595.84 v = sqrt[595.84] ----> sqrt[x] = square root of x v = 24.409 m/s So there final velocity as the rock hits the bottom of the hole is 24.409 m/s. Notice that the acceleration is now postive. Q. How long is the rock in the air from the it is released until it hits the bottom of the hole? This question is a lot easier to answer now that we know the peak height of the rock. To solve this the time taken for the rock to reach its peak height must be solved separtely to the time it takes for the rock to fall from its peak height to the bottom of the hole (then add them together). So first choose an appropriate equation. Im going to choose v = u + at as we have all the parameters apart from time which in this case we want to find. So from when the rock is thrown to when it reaches its peak height: v = u + at t = (v - u) / a t = (0 - 20) / -9.8 t = 2.040 seconds Now from the rocks peak height to the bottom of the hole: t = (v - u) / a t = (24.4 - 0) / 9.8 t = 2.49 seconds Therefore total time: t = 2.04 + 2.49 t = 4.53 seconds

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