shotputter

[logistic_guy]

A shotputter throws the shot (mass \(\displaystyle = 7.3 \ \text{kg}\)) with an initial speed of \(\displaystyle 14 \ \text{m/s}\) at \(\displaystyle 40^{\circ}\) angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlet's hand at a height of \(\displaystyle 2.2 \ \text{m}\) above the ground.

 

[logistic_guy]

There is a nice formula for an object moving horizontally with a constant velocity.

\(\displaystyle x = x_0 + v_0t\)

The problem of this formula is that we don't have the time to calculate the distance. Therefore, we have to think of something else!

 

[The Highlander]

There is a nice formula for an object moving horizontally with a constant velocity.

\(\displaystyle x = x_0 + v_0t\)​

The problem of this formula is that we don't have the time to calculate the distance. Therefore, we have to think of something else!

But, of course you can easily find the relevant time interval!

This is a straightforward (standard) Physics problem (in Newton's Equations of Motion, qv) that I've been teaching HS students how to do for thirty years.

The trajectory of the shot is a parabola (below is an accurate representation of it)...

​

When the shotputter throws the shot it is travelling up and forward at the same time. How long will gravity take to slow it down to zero (and at what height will this occur) before it then starts to accelerate back down towards the ground? And how long will it take to hit the ground?

Once you have those times you can use your simpler formula (above) to determine how far it has travelled horizontally.

 

[logistic_guy]

But, of course you can easily find the relevant time interval!

Of course I can. But you have to know that I don't look at any kind of help at the beginning of the attack. I depend fully on myself. I studied Physics a long time ago, so I forgot some of the formulas, but because I was very good in this field, I have so many ideas. My strategy to solve is that I try to apply my accumulated knowledge which helps me revive and derive the formulas again.

that I've been teaching HS students how to do for thirty years.

Thirty Years!you are just amazing.

I like your sketch. The same idea I had in mind. I just was lazy to draw it!

When the shotputter throws the shot it is travelling up and forward at the same time. How long will gravity take to slow it down to zero (and at what height will this occur) before it then starts to accelerate back down towards the ground? And how long will it take to hit the ground?

Now I have some other ideas to attack the problem. I will apply them first, if they fail, I will try your suggestion.

 

[The Highlander]

Now I have some other ideas to attack the problem. I will apply them first, if they fail, I will try your suggestion.

You've now had ample time to try out your "other ideas", are you going to complete this thread now you've started it or are you just going to leave it unfinished like so many of the others that you unreasonably begin (on stuff like bookkeeping or accountancy which have absolutely no place in this forum)?

 

[logistic_guy]

You've now had ample time to try out your "other ideas", are you going to complete this thread now you've started it or are you just going to leave it unfinished like so many of the others that you unreasonably begin (on stuff like bookkeeping or accountancy which have absolutely no place in this forum)?

My idea is to find the time directly by:

\(\displaystyle y = y_0 + v_0t + \frac{1}{2}gt^2\)

I am not sure if it works but let us try.

\(\displaystyle -2.2 = 0 + 14\sin 40^{\circ}t - (0.5)9.8t^2\)

This gives:

\(\displaystyle t = 2.1 \ \text{s}\)

Then, the horizontal distance is:

\(\displaystyle x = x_0 + v_0t = 0 + 14\cos 40^{\circ}(2.1) = 22.5 \ \text{m}\)


FYI. I forgot this formula \(\displaystyle y = y_0 + v_0t + \frac{1}{2}gt^2\). I had to derive it by calculus!