Question

An object is projected with a velocity U at an angle R to the horizontal.It just clears two walls of equal height b and at a distance of b apart. Show that 2U cosR√(U² sin²R - 2bg) = bg?

Answer 1

Explanations & Calculations

• If the vertical component of the velocity at the first time the trajectory marks its height b is $\small \uparrow V$ then by applying $\small v^2=u^2+2as$ for the vertical motion from start to that point we can link those moments.

$\qquad\qquad \begin{aligned} \small V^2&=\small (U\sin R)^2+2(-g)b\\ \small V&=\small \sqrt{U^2\sin^2 R-2gb} \end{aligned}$

• Then we can calculate the time it takes to travel between the two pillars of height b or the distance in between them; b. Make note the objects horizontal velocity is a constant throughout which is $\small U\cos R.$

$\qquad\qquad \begin{aligned} \small \to\\ \small s&=\small ut\\ \small b&=\small U\cos R.t\\ \small t&=\small \frac{b}{U\cos R} \end{aligned}$

• Then, during this time it travels vertically as well but makes a 0 vertical displacement as a whole.
• By this happening vertically,

$\qquad\qquad \begin{aligned} \small \uparrow\\ \small s&=\small ut+1/2at^2\\ \small 0&=\small Vt+\frac{1}{2}(-g)t^2\\ \small t&=\small \frac{2V}{g} \end{aligned}$

• Finally, equalling those two obtains via time and through a little bit of arrangement, we get what we need.

$\qquad\qquad \begin{aligned} \small \frac{2V}{g}&=\small \frac{b}{U\cos R}\\ \small 2U\cos R.V&=\small gb\\ \small bg&=\small 2U\cos R.\sqrt{U^2\sin^2 R-2gb} \end{aligned}$