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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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