Calculating Projectile Velocity in Russian Roulette: Bill Bored's Risky Game

[ccf788]

Bill Bored is out to get his kicks. He starts playing Russian roulette with a 3.25 kg shot put. He takes his chances throwing the ball straight up into the air and waiting to see if the ball hits him in the head on its way down. Bill is 1.95 m tall and the ball rises 4.0 m above the point of release (even with the top of his head).
a) How fast did the shot put leave Bill's hand?
b) How fast will the shot put hit Bill's head if he loses?
c) How fast will the shot put hit the ground if he wins and the ball misses him?

My problem is I don't know what equation to use. I was thinking to use 1/2mv^2 + mgy = 1/2mv^2 +mgy, but by asking how fast, are they asking for how fast in time or in velocity?

If they are asking for it in terms of velocity would my y1 be 1.95 m and my y2 be 5.95 m because it rises 4 m above release? For pt. a would my v1 be zero then?

Please also help me set up pt. b and pt. c. Thank you!

 

[dextercioby]

U could use the energy conservation law,but i find the method using the motion equation more direct.Since the movement is uniformly accelerated ("g" doesn't vary on 5.75),the equation of motion is (assume the Oy axis oriented upwards):
$$y(t)=y_{0}+v_{0}t-\frac{1}{2}gt^{2}$$(1)
For this kind of movement,the velocity is given by:
$$v_{y}(t)=v_{0}t-gt$$ (2)

These two equations are enough to lead you to the answers to points a),b) and c).

Daniel.

 

[wais]

a) To calculate the velocity at which the shot put left Bill's hand, we can use the equation v = √(2gh), where v is velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height at which the ball was released (4.0 m above Bill's head). Plugging in the values, we get v = √(2*9.8*4.0) = 8.86 m/s.

b) If Bill loses and the shot put hits him in the head, we can use the equation v = √(2gh) again, but this time h will be the total height the ball has fallen (4.0 m above Bill's head + 1.95 m, Bill's height). So, v = √(2*9.8*5.95) = 10.89 m/s.

c) If Bill wins and the shot put misses him, we can use the equation v = √(2gh) again, but this time h will be the height at which the ball hits the ground (4.0 m above Bill's head). So, v = √(2*9.8*4.0) = 8.86 m/s. This is the same velocity as in part a, since the ball is falling from the same height.

Note: In this context, when we say "how fast", we are referring to the velocity (speed and direction) of the shot put. So, for all three parts, we are calculating the velocity of the shot put.