Question

A skeet shooter aims at a clay pigeon. The muzzle velocity of the rifle is 700 mph and the man holds the rifle such that the end of the gun is 6 ft off the ground. Assuming that the speed of the skeet is insignificant relative to the speed of the bullet, determine the angle at which the man needs to hold the rifle in order to hit a clay pigeon that is 50 yards away and at an altitude of 30 ft.

Answer 1

so according to the question the bullet must travel through a distance of 50 yards

and climb a height of (30 - 6 ) = 24 feet

given that the velocity of the bullet = 700 mph = 1026.667 ft/s

let the angle be theta as shown in the figure

horizontal component of velocity = 1026.667cos(theta)

time taken to travel 50 yards (150 ft ) = 150 / [1026.667 cos(theta) ]

=0.146/cos(theta) seconds

so in 0.146/cos(theta) sec the bullet must climb a height of 24 feet

vertical component of velocity = 1026.667sin(theta)

acc.due to gravity = - 32 ft/s^2 (taking upward direction as positive)

using s=ut+0.5at^2

we have

24 = 1026.667sin(theta) *t -0.5*32*t^2

t = 0.146/cos(theta)

24 = 1026.667 sin(theta)*0.146/cos(theta) - 16(0.146/cos(theta))^2

24 = 150 tan(theta)-0.342 sec^2(theta)

using sec^2(theta) = tan^2(theta) +1

24 = 150 tan(theta) - 0.342[tan^2(theta) +1]

0.342tan^2(theta) - 150 tan(theta) +24.342 = 0

upon solving this quadratic equation we get

tan(theta) = 0.16234 , 438.434

giving theta = 9.22 degrees and 89.869 degrees