You throw a stone off of the edge of a high cliff. the stone leaves your hand go

Question

You throw a stone off of the edge of a high cliff. the stone leaves your hand going 20.0 degree above the horizontal at 22.0 m/s. The cliff is 15.0 m high If you are standing right at the edge of the cliff when you throw the stone and if the cliff is completely vertical, how far from the base of the cliff does the stone land? You have probably assumed in all of the above that air drag has no influence on the stone. Now suppose there is a steady wind towards the cliff, causing the stone to accelerate at 1.00 m/s2 horizontally during its flight. Does this change how much time the stone spends in the air? Explain Does this change the answer for b)? Explain. If it causes a change to either, recalculate using the ct of the wind in the calculation.

Explanation / Answer

Ux = 20.67 m/s

Uy = 7.522 m/s upwards

time t

-15 -7.522*t +4.9*t*t =0

t=2.678 sec

distance from base = 20.67*t = 55.35 m

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