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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)
 Feb 13, 2014
 #1
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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.
 Feb 13, 2014
 #2
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Thanks alot Rom!
 Feb 14, 2014

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