Question

A projectile is launched horizontally over a flat terrain from height h with an unknown initial speed. Just before the projectile lands, its velocity makes an angle θ with the horizontal. The acceleration due to gravity is g. What is its initial speed and its horizontal range?

Answer 1

This is pretty tedious as is, so I'm gonna assume you're somewhat familiar with kinematics equations for projectile motion: http://tatania.phsx.ku.edu/Phsx114/images/Kinematics.jpg
First find the time, which is simple since initial y-velocity is zero

\(\large h = \frac{ 1 }{ 2 } g t^2 \) gives \(\large t = \sqrt{\frac{ 2h }{ g }} \)

Now that we have t, we can find the final y-velocity, which remember from trig, is equal to vsinθ:

\(\Large v \sin \theta = g t \) or \(\Large v \sin \theta = g \sqrt{\frac{ 2h }{ g }} \) and simplify and solve for the final speed, v:
\[\Large v = \frac{ \sqrt{2hg} }{ \sin \theta } \]

Answer 2

Now we can find the initial velocity, \(\large v_x\) which is vcosθ, by plugging in that expression for v:

\(\Large v_x= \cos \theta \frac{ \sqrt{2hg} }{ \sin \theta }\) or \(\Large v_x = \cot \theta \sqrt{2hg} \)

Now it's easy to find the range, just use \(\large x = v_x t\) and plug in the expression for \(\large t\) from earlier, and \(\large v_x\)
\[\large x = \cot \theta \sqrt{2hg} * \sqrt{\frac{ 2h }{ g }}\]and simplfiy\[\large x = 2 h \cot \theta \]

Answer 3

But what about initial speed?

Answer 4

if it falls h, its vertical velocity is \(v_y = \sqrt{2gh}\) from energy conservation or the equations of motion, see agent smith's workings

its horizontal speed is a constant \(v_x\)

so \(\tan \theta = \dfrac{v_y}{v_x}\)

as per the drawing