You want to throw a 500 gram rock through a third story window in such a way tha

Question

You want to throw a 500 gram rock through a third story window in such a way that it travels horizontally through the window. You can throw the ball at 22 m/s and when you release the ball it is 10.0 m below the window. Show your reasoning. No credit for plug and chug into the range formula. If you do it the easy way, you will not need the trig 10: At what angle from the horizontal must you throw the ball? How far away from the building must you stand to throw it? What energy must you give the ball, relative to its release point? What kinetic energy does it have just before it hits the window? Rats! The rock doesn't break the window, it bounces off with half its impact speed. How much work did the window do on the rock? What speed does the rock hit the ground with?

Explanation / Answer

Its given stone hits window horizontally => V[y] is zero there & V[x] is 22cos m/s

Using 0 = V[y]2 - (2*g*10) => V[y] = 14.142 m/s

=> 22sin = 14.142 => = sin-1(14.142/22) = 40 degrees (ans)

time for V[y] to be zero = 14.142/10 = 1.4142 sec

Horizontal dist from building = 1.4142 * 22cos(40) = 1.4142 * 22 * 0.766 = 23.833 metres

The imparted energy is only the kinetic energy .e. KE = 1/2 * (0.5) * (222) = 121 Joules

Just before it hits the window, velocity = 22cos(40) = 16.853 m/s (impact velocity)

KE just before hitting window is 1/2 * (0.5) * (16.8532) = 71 Joules

Half the impact velocity = 16.853/2 = 8.4265 m/s

New KE = 17.75 Joules

The difference in KE is the work done by the window = 17.75 - 71 = -53.24 Joules (ans)

The V[y] returns to 14.142 m/s and V[x] remains constant at the new 8.4265 m/s

i.e. [14.1422 + 8.42652] = 16.462 m/s (ans) speed with which it hits ground

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