# Minimum speed required to clear rectangular object

I believe this comes under the "homework question" category, I will try to make it as general as possible to avoid concerns about cheating/extra help.

The problem is to find the minimum initial speed required to launch a projectile over rectangular object of width $x$ and height $y$.

My question is : is it a correct assumption to state that the tangent to the path of the velocity at the point where it passes the first corner of the rectangle is at 45° to the horizontal?

• Why do you think that assumption applies? What prevents the angle from having any other value? – sammy gerbil Oct 22 '16 at 18:20

My question is; is it a correct assumption to state that the tangent to the path of the velocity at the point where it passes the first corner of the rectangle is at 45° to the horizontal?

You're somewhat conflating things, as the figure below should make self-evident:

The optimum launch angle $\theta$ is 45° ($\frac{\pi}{4}$) but that's not the angle at which the projectile will 'clear' the first left hand corner.

At $\frac{\pi}{4}$ it becomes a matter of finding the intial velocity $v_0$ and the distance between the launchpoint and the mid point of the base of the object.

This Wiki entry should make calculation of these parameters quite easy.

• I have just been informed that my original assumption was correct, as you do not care about the total distance covered compared to the speed, but rather the lowest speed needed to travel the distance of the rooftop. Thank you kindly for helping me to achieve the wrong answer sir! – Mike D.H. Oct 29 '16 at 16:40

You are correct. The minimum speed $v$ needed to traverse the roof requires an angle of $45^{\circ}$ at each side of the roof. Applying the formula for maximum range equal to the width of the roof will give you the speed $v$ at each edge of the roof as a function of $x$ and $g$.

The speed $u$ required on launch from the ground is then easily found from conservation of energy : the KE lost in getting up to the edge of the roof equals the PE gained. The result is a very simple relation between $u,x,y$ and $g$.

• This is correct and exactly the solution that was discussed. Thank you. I wish I had read this answer before Gert's. – Mike D.H. Oct 29 '16 at 16:42