The Interrupted Projectile (that hits something on the way) Many projectiles don

Question

The Interrupted Projectile (that hits something on the way) Many projectiles don't complete their entire travel to the ground. We may need to calculate the range of a projectile that lands on a hill above ground or the height when the projectile stuck its mark. How would the flight time of a projectile that lands ABOVE the height it was launched compare to the flight time of a ground-to-ground projectile? ________ Why is this? ___________ What formula can we use that supports your explanation? ___________

Explanation / Answer

let the height abovethe projectile launchpoint be h
then, time of flight = t
h = u*sin(theta)*t - 0.5*g*t^2
here,, theta is angle of projectile and u is the initial projectile launch velocity
gt^2 - 2usin(theta)*t + 2h = 0
solving for t
a. t = [2u*sin(theta) +- sqroot(4u^2sin(theta) - 8hg)]/2
this is less thant the time period of a normal projectile that lands at the same height

b. This is because the prohectile coes to a stop before it actually has exhausted all of its potential energy to gainkinetic energy initially provided to it
c. t = [2u*sin(theta) + sqroot(4u^2sin^2(theta) - 8hg)]/2
for h = 0
T = [u*sin(theta)]

T - T' = [2u*sin(theta) + sqroot(4u^2sin^2(theta) - 8hg)]/2 - 2[u*sin(theta)]/2 = sqroot(4u^2sin^2(theta) - 8hg)/2 > 0

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