The vertical component of the velocity is 25sin20° = 8.55
So, the height y is
y = 18 + 8.55t - 16t^2
If the missile just clears the wall, then
40 = 18 + 8.55t - 16t^2
Hmm. No solutions.
Check for typos, and take a look at wikipedia's article on "trajectory" for details of the range, height, etc.
initial horizontal velocity: 25*cos(theta)=v1
initial vertical velocity:
where theta is the angle to the horizontal.
will it go over the wall:
solve the above equation for t and plug into:
18+v2*t-(32ft/s^2)/2*t^2= height at given time
(v2)^2-2*32ft/s^2*(height)=0" is wrong
it should be:
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