Projectile Problem: Solving for θ=60° | Sinθ = ±√3 Solution

[AakashPandita]

See the attachment.

The right answer is θ=60 degrees

but i am getting sinθ = ±√3.

 

[Saitama]

See the attachment.

The right answer is θ=60 degrees

but i am getting sinθ = ±√3.

Your expression for $v_y$ is wrong.

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

 

[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.

 

[AakashPandita]

First things first.

How is my expression for v_y wrong?

I used v²-u² = 2as

 

[Saitama]

First things first.

How is my expression for v_y wrong?

I used v²-u² = 2as

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

 

[haruspex]

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
$$mgH+ \frac{m u^{2} \cos^{2}( \theta)}{2}= \frac{mu^{2}}{2},$$

Doesn't that reduce to the same equation?

 

[Ackbach]

Doesn't that reduce to the same equation?

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.