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October 20, 2014

October 20, 2014

Posted by **Jessica** on Thursday, February 24, 2011 at 9:37pm.

- Physics 2(Plz Help! :) :) ;)) -
**MathMate**, Friday, February 25, 2011 at 4:00pmFirst we need to derive the horizontal distance covered for a muzzle velocity of u, and angle with the horizontal θ.

The time t the bullet stays in the air is

Sy=0 = u*sin(θ)*t - (1/2)gt²

Solving for t gives t=2u*sin(θ)/g.

Substitute t in the usual formula for horizontal distance:

Sx=(u*cos(θ))*t

=2u²sin(θ)cos(θ)/g

=u²sin(2θ)/g ....(1)

For the present case,

Sx = 10,000m

Solve for in equation (1)

10000 = u²sin(2*45°)/g

u=sqrt(10000g/1)=313.05 m/s

To travel less distance horizontally, we cannot change u, but we can vary θ such that:

5000 = 313.05&aup2;*sin(2θ)/g

or

sin(2θ)=5000g/313.05²=0.5

asin(0.5)=30° or 120°.

Therefore θ can be 15° or 60°.

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