A cowboy wishes to display his shooting skills to his friends. He asks a gentlem

Question

A cowboy wishes to display his shooting skills to his friends. He asks a gentleman, standing 8.0m from him, to toss a coin, and tells his friends that he can shoot the coin in the air. Having done his homework, he knows that the muzzle velocity of his Colt revolver is 73 m/s, and that the man will toss the coin upwards with a speed of 3.2 m/s.

The cowboy shoots at the same moment as the man tosses the coin, and the height above the ground of the revolver is the same as the mans hand, 1.0m

a) At what angle from the horizontal does the cowboy aim?
b) At what height above the ground does the bullet strike the coin?

Explanation / Answer

a) At what angle from the horizontal does the cowboy aim? let the cowboy aims at an angle ? above the horizontal .let the time taken be t and height be y. y=73sin?*t-1/2gt*t -----1 for bullet y=3.2t-1/2gt*t     ------2 for coin when they meet y is same for both. so equating these two equations we will get ?=25.990 regarding horizontal displacent it should be 8m since tossed coin will only move vertically up and down as it was projected vertically upwards. so 73cos25.99*t=8 so t=0.1219sec
b) At what height above the ground does the bullet strike the coin? regarding horizontal displacent it should be 8m since tossed coin will only move vertically up and down as it was projected vertically upwards. so 73cos25.99*t=8 so t=0.1219sec so y=3.2*0.1219-1/2*9.8*0.1219*.1219=0.318m

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