A carnival game requires the participant to toss a golf ball through a small ele

Question

A carnival game requires the participant to toss a golf ball through a small elevated ring (something like a scaled-down version of shooting a basketball free throw). The narrow hoop is elevated exactly 4.00 m above the participant's release point. In one particular winning toss, the ball descends with a total speed of 7.5 m/s at the instant is passes through the hoop, traveling at an angle of 50o from the horizontal. How much time elapses from the moment the ball is released until it passes through the hoop/ring? How far back (horizontally) is the participant from the hoop? What is the maximum height reached by the ball along its trajectory? What was the ball's initial velocity at the moment of release (state both the speed and the angle)? If the ball passes cleanly through the hoop with no collision, how far away (measured horizontally) will it be when it returns to the elevation from which it was released?

Explanation / Answer

speed of descent   = 7.5 m/s

angle of dsecent = 50 deg

horizontal speed uh = 7.5Cos(50) = 4.82 m/s

vertical speed = 7.5Sin(50) = 5.75 m/s

The ball has reachd maximum height and reachd 4 m height

if t is the time of descent form maximum height then

v= gt2/2   = 5.75

t2 = sqrt(2*5.75/9.8)

t = 1.08 s

height descended = u2/2g = 5.752 /2*9.8 = 1.69 m

time to reach ground   is given by

s = ut+gt2/2

4 = 7.5t+9.8t2/2

t = 0.42s

Total time of descent = 1.08 +0.42 = 1.5 s

                                    = time of ascent

A) time elasped   1.5 +1.08 = 2.58s

B) Horizontal distance taraveled before the loop = uh * t = 4.82*2.58 = 12.44 m

C) Maximum height reached = 1.69 + 4 = 5.69 m

                                               = total height descended

D) horizontal vel uh = 4.82 m/s const

time of acent = 1.5 s

intial vertical vel uv = gt = 9.8*1.5 = 14.7 m/s

inital vel (mag) = sqrt(14.72 + 4.822 ) = 15.47 m/s

angle = arcTan(15.47/4.82) = 72.69 deg with the horizontal

E) Total time of flight = 1.5*2 = 3s

Total horizontal disatcne (range ) = 3*4.82 = 14.46 m

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