Posted by **Lindsay** on Wednesday, November 7, 2007 at 9:09pm.

The launching speed of a certain projectile is 6.4 times the speed it has at its maximum height. Calculate the elevation angle at launching.

I have noo idea how to start this one. Please help!

- Physics -
**drwls**, Wednesday, November 7, 2007 at 11:41pm
The vertical component of velocity at max height is zero. The horizontal component remains Vh during flight, and that is the speed at max height. Let Voy be the initial vertical component of the launch speed.

(launch speed)^2 = Voy^2 + Vh^2

= (6.4 Vh)^2 = 40.96 Vh^2

Voy^2 = 39.96 Vh^2

Voy = 6.32 Vh

launch angle arctan(Voy/Vh) = arctan 6.32 = 81.0 degrees

- Physics -
**Lindsay**, Thursday, November 8, 2007 at 12:08am
How did you come up with 39.96 Vh^2?

- Physics -
**drwls**, Thursday, November 8, 2007 at 3:10am
I squared 6.4 and subracted 1. See the formula

Voy^2 + Vh^2 = 40.96 Vh^2

- Physics -
**Lindsay**, Thursday, November 8, 2007 at 7:58am
This is probably really obvious, but I'm still not undertanding why you have to subtract one.

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