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A certain projectile entry launches the projectile from the floor level. The target is placed on the wall such that the center of the target is 100cm above the floor. The ball is launched 4.2 m from the wall and at an angle of 60 degrees above the floor. The ball makes a direct hit dead center on the target. Calculate a] the time the ball is in the air. b] the resultant speed of the launched projectile. c] the component velocities of the ball from the moment of launch.

billrussell42 · Answer

Let V be the velocity at launch and Vx and Vy the components.

Vertically, there are two solutions. (1) the ball raises to a peak and falls back to the 1 m point, or (2) the ball is still raising at the 1 m point.

Horizontally, Vx = 4.2/t
Vx = V cos 60 = V/2 = 4.3/t
t = 4.3 / (V/2) = 8.6/V

Vertically, Vy = V sin 60 = V√3/2
case 1, time to peak t1 = Vy/g = √(2h/g)
h = Vy²/2g = (V√3/2)² / 2g = 3V²/8g
time to fall back to 1 m
t2 = √(2(h–1)/g)
t = t1+t2 = √(2h/g) + √(2(h–1)/g)
t = t1+t2 = √(2h/g) + √((2h–2)/g)
t = t1+t2 = √(2h/g) + √((2h/g) – (2/g))
t = t1+t2 = √(2(3V²/8g)/g) + √((2(3V²/8g)/g) – (2/g))
t = t1+t2 = √(6V²)/8g²) + √(6V²)/8g²) – (2/g))
t = t1+t2 = (V/3g)√(2) + √(6V²)/8g²) – (2/g))
8.6/V = (V/3g)√(2) + √(6V²)/8g²) – (2/g))
solve for V
this is getting complicated. Try case 2.

case 2
d = ½at² + v₀t
1 = –½gt² + Vt
gt² = 2Vt
t = 2V/g
8.6/V = 2V/9.8
2V² = 8.6•9.8
V = 6.49 m/s
you can do the rest.

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