A gunman standing on a sloping ground fires up the slope. The initial speed of the bullet is v0= 390 m/s. The slope has an angle α= 19 degrees from the horizontal, and the gun points at an angle θ from the horizontal. The gravitational acceleration is g=10 m/s2.
(a) For what value of θ ( where θ>α) does the gun have a maximal range along the slope? (in degrees, from the horizontal)
(b) What is the maximal range of the gun, lmax, along the slope? (in meters)
lmax=
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lmax= (v_0^2*(1-sin(alpha))/(g*cos^2(alpha))
a)
(pi/2)+(alpha/2)
I supposed that when bullet hits the slope, velocity on axis Y will be zero and it gives me one equation. I also have two equations, one for x axe(distance), and one for y axe(also distance, line perpendicular to the slope at x = cos Alpha * Lmax).
After I solved it, I obtain an answer for theta which is less than 45 degrees and more that Alpha. Can it be a correct answer? I know that if there is no slope, just horizontal ground, max distance will be if shout at 45 degree.. So I doubt in my answer
The angle should be less than 45 degrees
the angle should be more than 45°!
45° is for a horizontal plane.
so theta = 45 + (alpha/2)
The angle should be less than 45 degrees
how is that possible if 45+a/2 is the correct answer then the angle is more than 45 degrees..