# Projectiles - golf ball

Ok, having studied projectiles over the past few weeks, reaching the "Equation for the path of a projectile stage," I've been presented with the following question. As far as I can see, there isn't enough info to answer the question (unless, of course, its one of those simultaneous equation ones, or I've missed something obvious)

A golf ball is struck from a point O, leaving O with a speed of 21 root 2 m/s. The ball lands, without previously bouncing, at a point A, which is 10 metres below the horizontal plane through O. Find the speed of the ball at the instant it lands at A

In the question, g (Gravity) is taken to be equal to 9.8m/s^2...

Any help would be much appreciated!

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Doc Al
Mentor
There are several ways to solve this. If you've covered conservation of energy, use it.

Assuming you haven't, then considered the x and y components of the velocity separately. And realize that $V^2=V_x^2+V_y^2$.

Originally posted by Ihsahn
A golf ball is struck from a point O, leaving O with a speed of 21 root 2 m/s.

Excuse me, but I'm not familiar with the expression 21 root 2. What is it?

Originally posted by Julian Solos
Excuse me, but I'm not familiar with the expression 21 root 2. What is it?

Sorry if that was unclear, I meant 21 multiplied by the square root of 2.

Doc Al, I don't quite follow... we haven't covered conservation of energy... and I'm unsure how to obtain the x and y components given that we have no angle of projection? Forgive me if its something obvious or simple.

Thanks.

Doc Al
Mentor
Originally posted by Ihsahn
Doc Al, I don't quite follow... we haven't covered conservation of energy... and I'm unsure how to obtain the x and y components given that we have no angle of projection?
You don't have to know the angle. Call the components $V_x$ and $V_y$. Now write down the final values of those velocity components. Try it and see what you get. (Hint: Use $V_f^2=V_i^2+2as$.)

Again, sorry if this sounds stupid;

We've always evaluated (for example) $V_x$ as $V cos$ a and $V_y$ as $V sin$ a ??

The only thing I can think of from your suggestion is the use of trig identities? Do they play a part?

I'm obviously missing something...

Doc Al
Mentor
No trig needed. Do you agree that $V^2=V_x^2+V_y^2$?

So do the following...

1) Find $V_{y,f}^2$ in terms of $V_{y,i}$, $a$, and $\Delta y$ using the kinematic equation given earlier.

2) Realize that $V_x^2$ is a constant.

3) Add these to get $V_f^2$.

A solution (I think )follows. If you don't want to see it, ignore this post.

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Denote the initial velocity of the golf ball $$V_{0}$$ and the final velocity $$V$$.

We have a relation

$$V_{0}^2 = V_{x0}^2 + V_{y0}^2$$.

Substituting $$V_{0} = 21 \sqrt {2}$$,

we have

$$882 = V_{x0}^2 + V_{y0}^2$$.

Therefore,

$$V_{y0}^2 = 882 - V_{x0}^2$$.

Ignoring the effect of friction, the $$x$$ component of the speed of the golf ball remains unchanged throughout its flight.

Thus,

$$V_{x0} = V_{x}$$.

The $$y$$ component of the final speed of the golf ball upon landing is obtained from the following relation:

$$V_{y}^2 = V_{y0}^2 + 2g (10 \mbox{m})$$
$$= (882 - V_{x0}^2) + 196$$
$$= 1078 - V_{x0}^2$$
$$= 1078 - V_{x}^2$$.

$$V^2 = V_{x}^2 + V_{y}^2$$
$$= V_{x}^2 + (1078 - V_{x}^2)$$
$$= 1078$$.

$$V = \sqrt {1078}$$
$$= 32.83 \ \mbox{m/s}$$

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