Posted by Kayla on Tuesday, September 11, 2012 at 10:09pm.
A cannonball is shot (from ground level) with an initial horizontal velocity of 38.0 m/s and an initial vertical velocity of 26.0 m/s. The initial speed of the cannonball 46.04 m/s The initial angle θ of the cannonball with respect to the ground 34.38°
What is the maximum height the cannonball goes above the ground? (m)
How far from where it was shot will the cannonball land? (m)
What is the speed of the cannonball 2.7 seconds after it was shot? (m/s)
How high above the ground is the cannonball 2.7 seconds after it is shot? (m)

Physics  Steve, Wednesday, September 12, 2012 at 10:17am
vertical height h(t) = 26t  4.9t^2
max height reached at t = 26/9.8 = 2.653
h(2.653) = 34.49
Judging from parts 3 and 4, they expect the answer to part 1 to be 2.7 seconds.
In that case, v(2.7) = 0, h(2.7) = 34.5m
The range is given by
r = 46^2 sin 68.76° / 9.8 = 201.3m
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