# Help on Projectile Motion

1. Jul 31, 2007

### squall325

1. Angle = 15, Time(s) = 3.5, Horizontal Distance(m) = ?, Vertical Distance(m) = ?, Initial Velocity(m/s) = ?, Final Velocity(m/s) = ?

2. Our teachers told us to use y = vi(sine(theta)(t) - (gt^2)/2 - but we cant get an answer with only 2 givens.

3. Cant get an answer using the given equation. Hope you guys could help.

2. Jul 31, 2007

### Matthaeus_

Indeed, you can't solve for so many unknowns.

3. Jul 31, 2007

### Dick

Sure you can, if you assume 'time' is total time of flight and the surface is horizontal (etc etc). y=0 at t=0 and y=0 at t=3.5 sec. Use that to solve for vi and vertical distance. Now use x=vi*cos(theta)*t to do the horizontal part.

Last edited: Jul 31, 2007
4. Jul 31, 2007

### robsmith82

Maybe you are to assume the object is to be thrown from ground level? Then this is solvable.

5. Jul 31, 2007

### robsmith82

Maybe you are to assume the object is to be thrown from ground level? Then this is solvable.

6. Jul 31, 2007

### Matthaeus_

Ah, well, putting $$y_0 = 0$$ then you can solve:

$$\displaystyle \left \{ \begin{array}{ll} y &= (v_0 \sin 15^{\circ})t - \frac{1}{2} gt^2\\ v_y &= (v_0 \sin 15^{\circ}) - gt\\ x &= (v_0 \cos 15^{\circ})t \end{array}$$

7. Jul 31, 2007

### andrevdh

You can solve the parabolic equation

$$y = x \tan(\theta _o) - \frac{gx^2}{2v_o \cos^2(\theta _o)}$$

for the initial velocity $$v_o$$ - it is the only unknown in the equation if you use the other equations to subs for x and y.

But the problem do not require the initial velocity to be confined by any of the given values. This means that even if the projectile is resting on the launching pad at fifteen degrees and time marches on for 3.5 seconds the conditions are still met! So one can shoot the projectile at any initial speed and the situation will still be within limits of the given data.

Last edited: Aug 1, 2007