Two Physics students have identical projectile launchers that launch the ball wi

Question

Two Physics students have identical projectile launchers that launch the ball with the same initial speed. The projectile launchers are positioned so that the upper one is directly above the lower projectile launcherand separated by a vertical distance. The upper launcher fires the ball horizontally. The lower launcher is set up to fire at an angle of 53.13 degrees above horizontal giving the ball a velocity of 48 m/second at the highest point of point of its flight. The balls from both projectile launchers hit the same target that is at the same elevation of the lower projectile launcher. Air resistance is negligible. The time that the ball fire from the lower projectile launcher is in the air is? The time that the ball fired from the upper launcher is in the air is? The horizontal distance to the target from the release points of the projectile launcher is? The distance (Y) that the upper launcher is placed directly above the lower launcher is?

Explanation / Answer

given that u*cos(theta) = 48 m/s

u = 48/cos(53.13) = 80 m/s

R1 = R2

u*T1 = u*cos(theta)*T2

T2/T1 = 1/cos(53.13) = 1.66

R2 = u^2*sin(2*53.13)/g = 626.3 m = u*cos(theta)*T2

T2 = 626.3/48 = 13.04 S.......time that ball fired from the lower launcher

u*T1 = 626.3

T1 = 626.3/80= 7.82 S.....time that ball fired from the upper launcher

from the upper launcher R1 = 626.3 m

T1 = sqrt(2h/g)
7.82*7.82 = 2h/g

h = 299.95 m = 300 m

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