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


1 Answer 1


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 $sin(\theta)$.


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