Grade 12 PHYSICS Homework Help?

Question

Not too long ago, an interesting thing happened at the Canadian Open Golf Tournament. Tiger Woods bounced his tee shot off of a young spectator’s head in the last round of the tournament. The ball made contact for 0.08 seconds, and struck the boy’s head at a velocity of 32 m/s, 25 degrees from a tangent on the surface of his cranium. It bounced off at an angle of 22 degrees, with a speed of 25 m/s. What was the average acceleration of the ball as it amazingly made its way back onto the fairway?

Please try to solve this, and explain as clearly as you can! I don't wanna answer something without knowing what I am doing first :)

Equations we used up to this point were the 5 basic kinematic equations plus these 2:

Delta (d) h = [(V1)^2sin2@]/g
dt = [2V1sin@]/g

Thanks!

Answer 1

The equation to use here is:

average acceleration = (the change in velocity)/(time taken for the change in velocity)



We know:

Time taken for the change in velocity = time of contact with the poor dude's head = 0.08s



Here's the trickier part: we need to find the change in velocity. Now, velocity is a vector and we need to find the difference between the final velocity vector and the initial velocity vector. I am assuming you know a bit of vector math. Here's the link to a crude plot I made to help you follow the stuff written below: http://imgur.com/v8TyUcS



Initial vector is 32m/s at an angle of 25 degrees downward with the horizontal assuming that the tangent to the surface of the poor dude's cranium is perfectly horizontal. Let's call it -25 degrees

Final vector is 25 m/s at an angle of 22 degrees upward with the horizontal.

The vector difference = (final vector - initial vector) = 25 m/s (at 22 degrees) - 32 m/s (at -25 degrees) = 23.6 m/s (at 104.3 degrees)



So, the magnitude of average acceleration is 23.6/0.08 = 295 m/s^2 and the direction of the average acceleration = 104.3 degrees



Thanks and don't forget to vote if you liked the answer ;)

Answer 2

If U want a clear explanation, then ditto for the description :>}

This is one of those physics questions where a diagram/picture would be in order.

The question left "open to assumption" is the relative directions of the ball.



I shall assume ball strikes the cranium and bounces off the cranium in the SAME relative direction.

The method of solution is based on:

Impulse = change in momentum

Favg • t = mΔV

ΔV = (Favg • t)/m = (avg a) • t

avg a = ΔV/t = ΔV/0.08



Vi = 32 m/s at 25°

Vix = 32(cos 25°) = 29.0 m/s

Viy = 32(sin 25°) = 13.5 m/s



Vo = 25 m/s at 22°

Vox = 25(cos 22°) = 23.2 m/s

Voy = 25(sin 22°) = 9.37 m/s



ΔVx = Vox - Vix = 23.2 - 29.0 = -5.8 m/s

ΔVy = Voy - Viy = 9.37 - 13.5 = -4.13 m/s

ΔV = √[(-5.8)²+(-4.13)²] = √50.6969 = 7.12 m/s



avg a = 7.12/0.08 = 89.0 m/s² ANS*



*comment: THIS answer could have been approximated right away {that is, assuming that the relative directions before and after hitting by are nearly the same}. So because 25° and 22° are only 3° different ASSUME they are the same angle and thus subtract the two velocities to find ΔV ≈ 32-25 = 7 m/s and 7/0.08 ≈ 88 m/s <= approx answer even w/o the components being computed.

Answer 3

Δvx=v2cosθ2-v1cosθ1

Δvx=25*cos22-32*cos25=-5.82 m/s

Δv=v2sinθ2-v1sinθ1

Δvy=25*sin22-32*sin25=-4.16 m/s

Δv=(Δvx^2+Δvy^2)^1/2=6.64 m/s

a=ΔV/Δt

a=6.64/0.08=83 m/s^2