# 2D motion

1. Jun 10, 2012

### Kawakaze

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

A rugby ball of mass 400 grams is to be kicked from ground level to clear
a crossbar 3m high. The goal line is 12m from the ball, and the junior
rugby player can impart a maximum speed of 4√g ms−1 to the ball.

(a) Modelling the ball as a projectile moving under gravity alone, what is
the minimum launch angle that will succeed? What is the range of
the ball when launched at this angle and speed?

3. The attempt at a solution

Not quite sure how to approach this one at all. The ball would need to go high enough, from the initial vertical speed given to it - 4√g.sinθ. That by the time it reaches the crossbar in a time we can calculate from its horizontal component, it is exactly 3m above the ground. I get the feeling I am missing something though.

2. Jun 10, 2012

### Muphrid

I would find the time it takes for the ball to reach the crossbar as a function of angle. Then, you should be able to find an expression for the height of the ball at that time and see what constraints there are on the angle.

3. Jun 14, 2012

### Kawakaze

Thanks for the reply Muphrid. I made some progress thanks to your suggestion. I have an expression for the position vector. First problem is I dont know if they want a numerical or algebraic answer.

r=4t√g cos(α)i + (4t√g sin(α)-1/2 gt2)j + C

I need to get find the integration constant first, the only values i have are the initial values of (0,0) at time=0, giving C as zero, i dont know if this is correct.

As I need to find the minimum angle I assume that the position vector to the top of the wall is key to this (12i, 3j). Now I am stuck again.

4. Jun 14, 2012

### Muphrid

You might want to check the problem statement or something; you said that the player can impart a velocity of "$4 \sqrt g$" in meters per second, yet if $g$ is Earth's gravitational acceleration, that makes no sense in terms of units. Until that is cleared up, nothing else about the problem will make sense.

Otherwise, yes, you've reasoned out the integration constant correctly. The easiest thing to do is to solve for $\cos \alpha$ and then use inverse trig to get $\alpha$ from that.

Last edited: Jun 14, 2012
5. Jun 14, 2012

### Villyer

I would assume that it just means $4\sqrt{9.8}\frac{m}{s^{2}}$.

6. Jun 14, 2012

### azizlwl

2 constrains at time=t
1. x=12m
2.y≥3m or y=(3+z)m where z≥0

t=x/Ux

y+z=Uyt - 0.5gt2

minimum value of z=0 imply minimum angle.

3=Uy(x/Ux) - 0.5g(x/Ux)2

7. Jun 17, 2012

### Kawakaze

Azizlwl, thanks for the reply, but can you explain it a little more in depth please, i dont understand what your symbols mean, especially the z. Is U initial velocity?

8. Jun 17, 2012

### azizlwl

Yes U is the intial velocity.
z is a distance above the crossbar.
Ux initial horizontal velocity
Uy initial vertical velocity

9. Jun 17, 2012

### Kawakaze

Thanks! One last question if I may, what is the significance of the integration constant that appears when integrating the velocity vector to get the position vector, and where does it appear in your solution. Cheers!

10. Jun 17, 2012

y0

y=y0+ut+at2