Use the model for projectile motion, assuming there is no air resistance. The qu

Question

Use the model for projectile motion, assuming there is no air resistance. The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by a receiver 20 yards directly downfield at a height of 3 feet. The pass is released at an angle of 35 degree with the horizontal. (a) Find the speed of the football when it is released. (b) Find the maximum height of the football. (c) Find the time the receiver has to reach the proper position after the quarterback releases the ball.

Explanation / Answer

let v be the initial velocity
vy = vsin
vx = vcos

in the horizontal direction without friction
d = vx(t)
t = d/vx
t = d/vcos

if the origin is set at the ground level for vertical direction with up positive
y = y0 + v0t + ½gt²

3 = 6 + tvsin + ½(-32.2)t²
sub for t

3 = 6 + (d/vcos)vsin + ½(-32.2)(d/vcos)²
0 = 3 + dtan - 16.1d²/v²cos²
16.1d²/v²cos² = 3 + dtan
16.1d²/(3 + dtan)cos² = v²
v = [16.1d²/(3 + dtan)cos² ]

d = 20(3) = 60 ft

v = [16.1(60)²/(3 + 60tan35)cos²35 ]
v = 43.6697 ft/s ===> ANS a)

In the vertical plane
vf² = vi² + 2g(h)
at the apex of flight
0² = (43.6697sin35)² - 2(32.2)(h)
(h) = 9.80309 ft
add this to the throwing height to get
9.80309 + 6 = 15.80309 ft ...Answer B

t = d/vcos
t= 60/(43.6697*cos(35)) = 1.67728 sec

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