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September 16, 2014

September 16, 2014

Posted by **Peter** on Sunday, June 2, 2013 at 6:23pm.

If the missle is fired at a 35 degree angle, how high does it go?

- pre-calculus -
**Steve**, Monday, June 3, 2013 at 11:46amThe vertical component of the velocity is 25sin20° = 8.55

So, the height y is

y = 18 + 8.55t - 16t^2

If the missile just clears the wall, then

40 = 18 + 8.55t - 16t^2

Hmm. No solutions.

Check for typos, and take a look at wikipedia's article on "trajectory" for details of the range, height, etc.

- pre-calculus -
**Jay**, Wednesday, June 5, 2013 at 7:17pminitial horizontal velocity: 25*cos(theta)=v1

initial vertical velocity:

25*sin(theta)=v2

where theta is the angle to the horizontal.

max height:

(v2)^2-2*32ft/s^2*(height)=0

will it go over the wall:

v1*t=500 ft

solve the above equation for t and plug into:

18+v2*t-(32ft/s^2)/2*t^2= height at given time

- pre-calculus -
**Jay**, Wednesday, June 5, 2013 at 7:20pmone correction:

"max height:

(v2)^2-2*32ft/s^2*(height)=0" is wrong

it should be:

(v2)^2-2*32ft/s^2*(height -18)=0

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