# Problem on projectile

1. Jul 10, 2013

### AakashPandita

See the attachment.

The right answer is θ=60 degrees

but i am getting sinθ = ±√3.

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2. Jul 10, 2013

### Saitama

Your expression for $v_y$ is wrong.

Also, check your last step when you plug 3 in 4.

Last edited: Jul 10, 2013
3. Jul 10, 2013

### Ackbach

I don't agree with your very first equation. It is not true that
$$H= \frac{u^{2} \sin^{2}( \theta)}{2g}.$$
Conservation of Energy requires, instead, that
$$mgH+ \frac{m u^{2} \cos^{2}( \theta)}{2}= \frac{mu^{2}}{2},$$
comparing the peak to the starting-point, or
$$gH+ \frac{u^{2} \cos^{2}( \theta)}{2}= \frac{u^{2}}{2}.$$
Then you can also write
$$\frac{gH}{2}+ \frac{5 u^{2} \cos^{2}( \theta)}{4}= \frac{u^{2}}{2},$$
comparing the energies at the half-way point to the starting-point.

4. Jul 10, 2013

### AakashPandita

First things first.

How is my expression for vy wrong?

I used v2-u2 = 2as

5. Jul 10, 2013

### Saitama

What are the directions for velocity in vertical direction and acceleration?

6. Jul 10, 2013

### haruspex

Doesn't that reduce to the same equation?

7. Jul 10, 2013

### Ackbach

You're quite right. Chalk that one up to needing to do one step at a time. Also, for the half-way point, it doesn't reduce the same way.