Question

A rifle that shoots bullets at 471 m/s is to be aimed at a target 46.4 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Answer 1

Answer:

The equation that will express this result os

h = 0 = vy t - 1/2 g t^2    so the net height traveled by the bullet is zero

vy t = 1/2 g t^2

vy = 1/2 g t

vy = 1/2 * 9.8 * t       you could use -9.8 to indicate vy and g are in different directions

tx = sx/ vx  = 46.4 / 471 = .0985 sec    time to travel up and down to original height

th = .0985 / 2 = .0493 sec        time to reach maximum height

vy = g ty = 9.8 * .0493 sec = .483 m/s    initial vertical speed

Sy = vy t - 1/2 g t^2 = .483 * .0493 - 1.2 9.8 (.0493^^2)

Sy = .0238 - 4.9 ( .0493)^2 = .0238 - .0119 = .0119 m

Height to which bullet will rise - if the gun is aimed at this height then in .0985 seconds the bullet will fall to zero height

Check:    .483 / 9.8 =  .0493   time to reach zero vertical speed

total travel time = 2 * .0493 = .0986 sec

471 * .0986 = 46.4 m   total distance traveled by bullet