Projectiles

Quote:

A projectile is fired with initial velocity 5umeters/second at an angle of projection tan inverse of (4/3).
The projectile is halfway below its maximum height at times t1 and t2.
Show that:
a) t1 + t2 = (8u/g)
b) t1* t2 = (8u^2/g^2)

ok first off we note that the angle of projection is the midsized angle of a 3/4/5 triangle and thus it's cosine is 3/5 and it's sine is 4/5

so the projectiles initial vertical velocity is 5u*4/5 = 4u m/s

we can find it's maximum height by cheating a bit. The projectile has kinetic energy due to it's vertical component of (1/2)mv _vert ² = (1/2)16u ²m = 8u ²m
we know at the projectiles maximum height the potential energy due to gravity is equal to this kinetic energy. The potential at height h is mgh where g is the gravitational acceleration.
so the max height is at 8u ²m = mgh, h = 8u ²/g

You want to find the two times t1 and t2 where the height is half of this i.e. h(t1) = h(t2) = 4u ²/g

h(t) = (4u)t - g t ²/2

so just solve (4u)t - g t ²/2 = 4u ²/g

and this will get you your t1 and t2. Then just add them and multiply them and show that (a) and (b) are true.

Thanks alot Rom!