Math
posted by Devon .
A projectile is fired with an initial velocity of 350 feet per second at an angle of 45° with the horizontal. To the nearest foot, find the maximum altitude of the projectile. The parametric equations for the path of the projectile are
x = (350 cos 45°)t, and
y = (350 sin 45°)t  16t2.

Maximum height is achieved when
Vy = (Vo cos 45)*t g t = 0
t = (350 sin 45)/g = 7.69 s
where g = 32.2 ft/s^2
The height at that time is
Vo sin 45 t  (g/2) t^2
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