a shell was fixed horizontally at 325 m/s from a tower 175m high,what is time when hits the ground?

Answer 1

For any projectile fired at speed (u) with (angle p), position components are
x = u cos p * t
y = u sin p * t - 0.5 g t^2
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here p = 0 (horizontally), u = 325 m/s.
whwn it hits the ground >> y = - h = - 175 m
- 175 = 0 - 0.5 gt^2
gt^2 = 175*2 = 350
t^2 = 350/9.8
t = 5.98 sec (-ve time ignored) = time of flight
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distance travelled by shell (Range)
R = u cos p* t
R = 325 cos 0 * 5.98 = 1943.5 meters

Answer 2

Theoretically they're going to it the floor collectively; the vertical part of air resistance is the comparable no rely if that's a function of vertical velocity. yet air resistance isn't that straightforward. that's non-linear, has vortices, and modeling issues. The severe horizontal air resistance would carry over to the vertical and so the 2nd ball would hit first. i do no longer comprehend. Why do no longer you go up a 175m tower with a cannon and a pair of balls and let us know the end result?

Answer 3

I will let you do the math. acceleration due to gravity is 9.8 M/s/s. The fact that the shell is moving horizontily has no effect on how long it takes to fall to the ground. It still falls at 9.8 M/s/s. Oh I know that it will take a little longer to hit the ground due to curvature of the earth and therefore the shell will fall a little more than 175 meters.