Finding the Final Speed of a Golf Ball Projected at an Angle

[Ihsahn]

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!

 

[Doc Al]

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 $V^2=V_x^2+V_y^2$.

 

[Julian Solos]

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?

 

[Ihsahn]

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]

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 $V_f^2=V_i^2+2as$.)

 

[Ihsahn]

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]

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 $V_f^2$.

 

[Julian Solos]

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:

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



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


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