What Was the Initial Speed of the Shot Put in Randy Barnes' World Record Throw?

[godkills]

The men's world record for the shot put, 23.12m, was set by Randy Barnes of the United States on May 20, 1999. If the shot was launched from 6 ft above the ground at an initial angle of 42 degrees, what was its initial speed?

Known Data Is Height, angle of where it was shot, and how far it will land.

Thinking about using R = (Vo²/g)sin2θ but if only i can modify it so I can use it

anyone can help? suggestions? answer is 14.5 m/s from the back of book

this is from the physics book by james walker.

probably one of the annoying things I don't like is not knowing how to do it but knowing how to approach it -_-

 

[gneill]

I think you'll have to bite the bullet and write the appropriate equations of motion and solve for the initial velocity; essentially that's how you modify the range equation!

 

[godkills]

ughh i don't think i have to but isn't there some way to modify the range equation so i can find the initial velocity?

 

[Dickfore]

ughh i don't think i have to but isn't there some way to modify the range equation so i can find the initial velocity?

The equations of motion are:

$$x = v_{0} \, \cos{\theta} \, t$$

$$y = v_{0} \, \sin{\theta} \, t - \frac{g \, t^{2}}{2}$$

assuming that the initial position of the projectile (at $t = 0$) is the origin, the y-axis is directed opposite of the acceleration of free fall and the initial speed is $v_{0}$ and builds an angle with the horizontal is $\theta$.

One can get an equation for the trajectory by eliminating t. Solve the first equation for t:

$$t = \frac{x}{v_{0} \, \cos{\theta}}$$

and substitute in the second equation. We get:

$$y = v_{0} \, \sin{\theta} \, \frac{x}{v_{0} \, \cos{\theta}} - \frac{g}{2} \left(\frac{x}{v_{0} \, \cos{\theta}}\right)^{2}$$

This can be simplified with the use of the trigonometric identity:

$$\frac{1}{\cos^{2}{\theta}} = 1 + \tan^{2}{\theta}$$

to:

$$y = x \, \tan{\theta} - \frac{g \, x^{2}}{2 \, v^{2}_{0}} \, (1 + \tan^{2}{\theta})$$

This equation involves 4 quantities: the final position coordinates $(x, y)$, the initial speed $v_{0}$ and the angle of the initial velocity vector with the horizontal $\theta$ (assuming acceleration of free fall $g$ is known). If you know 3 of them, as in your problem you can solve for the fourth one (it is a quadratic equation though).

 

[ashishsinghal]

Why are you going in such a manner? Just use basic equations of motion, that is all

 

[godkills]

interesting, but i have no yet learned using the trigonometry in physics. I only know pythagoream therom tan cost but not the ones that you mentioned.