Physics 2(Plz Help! :) :) ;))
posted by Jessica .
A bullet leaving a rifle at an angle of 45 degrees travels a distance of 10 Km determine: A). The muzzle velocity B). What angle would you have to shoot the same bullet to go HALF the distance? Lower angle

First 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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