How Fast Was Randy Barnes' Record-Setting Shot Putt?

[leviathanX777]

Q1. A world record for the shot putt of 23.12 m was set by Randy Barnes in Los Angeles in 1990. Estimate the speed at which the shot had to be projected to reach this distance, assuming that the shot was released at 45 degrees to the horizontal from a height of 2.2m above the ground. (The ground is horizontal).

Q2. If the mass of the shot in part A was 2kg, find the total work done on it by Barnes in picking it up and projecting it, assuming that it was at ground level before he picked it up.



2. h=1/2gt^2 S=ut + 1/2 at^2



3. I resolved the vectors into the x and y coordinates but I'm not sure what I'm doing wrong as the back of the book has an answer of 14 m/s for the first part.

Thanks anyways!

 

[diazona]

Did you combine the x and y components of the velocity vector and find its magnitude?

Also, it'd be nice to show your work ;-) it's going to be hard to help you otherwise.

 

[nlightner]

Dear student,

I would be happy to assist you with your questions. Let's start with the first one.

To estimate the speed at which the shot had to be projected, we can use the equation for projectile motion:

S = ut + 1/2 at^2

Where:
S = distance traveled (23.12 m)
u = initial velocity (what we are looking for)
a = acceleration due to gravity (9.8 m/s^2)
t = time (unknown)

Since the shot was released at 45 degrees to the horizontal, we can resolve the initial velocity into its x and y components. The initial velocity in the y direction (vertical) would be usin45 = ucos45 = 0.707u. The initial velocity in the x direction (horizontal) would be ucos45 = 0.707u.

Now, we can use the equation for vertical motion to find the time it takes for the shot to reach its maximum height (when it is at 2.2m above the ground).

h = 1/2 gt^2

Where:
h = height (2.2m)
g = acceleration due to gravity (9.8 m/s^2)
t = time (what we are looking for)

Solving for t, we get t = 0.664 seconds.

Now, we can use this time to find the initial velocity in the x direction, since the horizontal motion is constant (no acceleration).

S = ut
23.12 = 0.707u * 0.664
u = 48.90 m/s

Therefore, the initial velocity of the shot must be 48.90 m/s for it to reach a distance of 23.12 m, assuming it was released at 45 degrees from a height of 2.2m.

For the second question, we can calculate the work done on the shot by using the formula:

W = F * d

Where:
W = work done (what we are looking for)
F = force (which is equal to the weight of the shot, which is mg)
d = distance (23.12 m)

Therefore, the work done by Barnes in picking up and projecting the shot would be:

W = mg * d
= 2 kg * 9.8 m/s^2 * 23.12 m
= 452.96 Joules

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