# Difficult projectile motion question

1. Jul 6, 2013

### sorax123

1. The problem statement, all variables and given/known data

A cricketer strikes the ball at a height of 1 metre. It passes over a fielder 7 metres from the ball at a height of 3 metres, and hits the ground 60 metres from the bat. How fast was the ball hit?

The issue I have here is that after forming a quadratic in terms of tanα, you have u in as an unknown and subbing it back in gives incredibly messy algebra, which makes me think it's not the best way.

2. Relevant equations

u in x direction= ucosα s in x direction= utcosα
u in y direction= usinα s in y direction= utsinα - 4.9t^2

3. The attempt at a solution

My solution attempt is pages long and contains stupidly complex algebra, I think I am fundamentally missing something as the maths seems too fiddly for a fairly straightforward mechanics course.
Any tips to make the problem easier, or preferably a solution would be great. Thanks

2. Jul 6, 2013

### SteamKing

Staff Emeritus
You have probably made an invalid assumption or a mistake in your algebra or arithmetic. Without looking at your calculations, it is hard to point to the specific cause of your difficulty.

3. Jul 6, 2013

### voko

Using the horizontal and vertical components of initial velocity, not the angle, the solution seems straightforward.

4. Jul 7, 2013

### verty

This is the important formula to know:

$y = y_0 + tan{θ} x - \frac{g}{2 v_0^2 cos^2{θ}} x^2$

It has the form

$y = c + bx + ax^2$

and the problem gives us 3 points, so apply Gaussian elimination (sorry, GE is not something you will have learned yet, you need to solve 3 equations in 3 unknowns) to find a,b,c.

PS. I think this formula is worth memorizing. You should be able to see where it comes from.

Last edited: Jul 7, 2013
5. Jul 8, 2013

### pgardn

If while passing over the fielder the projectile was at its max. Height it would make the problem more appropriate for the course level you describe. Did the problem perhaps say this? If the problem did state this you could solve for time in the air at that point and then everything becomes very straightforward.

6. Jul 8, 2013

### Staff: Mentor

Probably not a good assumption considering that the trajectory (a section of a parabola) must by symmetric about the point of maximum height. The fielder is only 7m from the batter. The ball lands 60m away.

One way to reduce the math burden might be to leave the initial vertical and horizontal components as separate variables, then write equations for the separate known points that the ball passes through. Hint: Lot's of nasty stuff should cancel cleanly if you translate the heights so that the bat is at height 0 (so, for example, the ball hits the ground at a "height" of -1m). Equate expressions for Vy to find Vx. This is one of those occasions where it's probably beneficial to substitute in the given numbers just before the end, allowing you to collapse a lot of terms before you hit the quadratic formula.

7. Jul 8, 2013

### pgardn

Yep.

My additional information makes no sense. And since they give them 3 points in a parabola...
It's a math problem.

Last edited: Jul 8, 2013
8. Jul 8, 2013

### voko

I would use the velocity components and exclude time, thus getting an equation for x and y. Then having a non-zero initial height should not complicate much of anything.

9. Jul 9, 2013

### rl.bhat

You can rewrite the general equation of the projectile
y = yo + x tanθ - 1/2*g*x2/v2cos2θ as

1/2*g*x2/v2cos2θ = yo + x tanθ - y.
Write down two equations for two values of x and y. By making the division, you can simplify the equations and then solve for tanθ.

Last edited: Jul 9, 2013