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Projectile Velocity and Vectors


Date: 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/   
    
Associated Topics:
High School Calculus

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