Post a New Question


posted by .

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.

  • Math -

    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

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question