# Math

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.

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1. 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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posted by drwls

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