Projectile Velocity and VectorsDate: 5/7/96 at 13:22:29 From: Anonymous Subject: Projectile velocity and vectors If the initial velocity is 834 meter^2/sec at an angle of 60 degrees, and x is 21,000, at what time, t, is the object at x = 21000 meters? I need the formula. Thanks. Date: 7/17/96 at 14:48:27 From: Doctor Robert Subject: Re: Projectile velocity and vectors If v is the initial speed in m/sec and A is the angle of inclination of a projectile, then the equations which give x and y as a function of time are x = (v cosA)*t and y = -.5gt^2 + (v sinA)t where g is the acceleration due to gravity. It has a value of approximately 9.8 m/sec^2. Substituting x = 21000 into the first equation we get t = 50.36. Substituting t=50.36 into the equation for y gives y = 23,951 m. This is the altitude when the object has traveled a HORIZONTAL distance of 21000 meters. Best of luck! -Doctor Robert, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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