Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I stack a question about projectile question.

The question was

A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height $h$ above the ground (a) what us the maximum launch speed you could give this projectile if you shot it straight up ? Express your answer in terms of $h$ and $g$.(b)Suppose the launcher available shoots projectiles at twice the maximum launch speed you found in part (a). At what maximum angle above the horizontal should you launch the projectile ?

I could solve the (a) part. How did was following (a) using the following formula to drive $V$

$\delta x$ = $\frac{V^{2}-Vi^{2}}{2g}$$

also we have $Vi = 0$, $\delta x = h$

I got $V = \sqrt{2gh}$

after that I think I need to use some kind of angle relative formula to create $arccosx$ or $arcsinx$ will be equal to some number then find the angle but I still don't have idea which formula do I need to use and find max angle.

also Do I need to split $Vx$ and $Vy$ from $V$ ?

One more question, I have seen some minimum spe

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

Part (b) is easy because you just need the vertical component of the velocity to be $\sqrt{2gh}$.

If you launch the projectile at an angle $\theta$ and velocity $v$, the vertical component of the velocity, $v_y$ is:

$$v_y = v sin(\theta)$$

You're told the projectile is launched at twice the velocity from part (a) i.e. $2\sqrt{2gh}$ so in the equation above set v to $2\sqrt{2gh}$ and $v_y$ to $\sqrt{2gh}$ and solve for $sin(\theta)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.