The drawing shows an exaggerated view of a rifle that has been \"sighted in\" fo

Question

The drawing shows an exaggerated view of a rifle that has been "sighted in" for a 91.5-meter target. If the muzzle speed of the bullet is v0 = 428 m/s, what are the two possible angles ?1 and ?2 between the rifle barrel and the horizontal such that the bullet will hit the target? One of these angles is so large that it is never used in target shooting. (Hint: The following trigonometric identity may be useful: 2 sin ? cos ? = sin 2?.)
?1 = ° (smaller angle)
?2 = ° (larger angle)

Please explain answers and formula, thanks.

Explanation / Answer

given that horizontal range R=91.5m    intial speed of the bullet is v_0=42.8m/s then R=(v_o^2/g)sin2? putting the values we get ?1=14.654deg                   and ?2= 90-14.654=75.345 deg

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