At impact, what is its speed? Answer in units of m/s.
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 12.6 m/s. The cliff is 24.4 m above a flat, horizontal beach. The acceleration of gravity is 9.8 m/s^2.
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What you have is a two dimensional motion question. Since there is no acceleration in the horizontal plane, that velocity would remain the same (constant). So you only have to solve for the vertical velocity at impact.
In solving this question you would use the equation d = Vit + 1/2at^2. Where the initial velocity multiplied by the time is added to half the acceleration times the time squared. You of course must solve this equation for t, and since Vi is zero, solving it is pretty easy.
THis will give you a flight time of 2.231499907, which you then use in the equation Vf = Vi + at, again Vi is zero and Vf is the impact velocity in the vertical plane. This should give you a vertical velocity of -21.86 m/s (the negative is important as it indicates direction). This is probably the answer they are looking for.
If the question was looking for the two dimentional impact speed, you would then use pythagorean theorem, with -48.8 and 12.6 being sides of a right triangle, which makes the hyptenus the two dimensional impact speed. Which would be 50.4 m/s.
Horizontal velocity of the stone = 12.6m/s
The height of the cliff = 24.4m
The initial vertical velocity of the stone = 0
Let the vertical final velocity be v m/s and initial vertical velocity be u and the displacement be s. Then they are related by the equation of motion:v^2-u^2= 2gs , where g is the acceleration due to gravity,
v^2-0^2= 2*9.8*24.4 or
v= sqrt(2*9.8*24.4) = sqrt(478.24) =21.8687 m/s
Therefore, the magnitude of the velocity (or speed) of the stone. = sqrt(horizontal velocity^2+vertical velocity^2)= sqrt(12.6^2+21.8687^2)=25.2389 m/s i
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