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M.Samad-CNAQ-
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Homework Statement
A projectile is to be launched by a marksman through an opening in a wall which is 40 m
away over a flat ground. The lowest part of the opening on the wall is 30 m above the ground
and the opening’s gap is 10m high. The mechanism for launching the
projectile generates an initial release velocity of 35 m/s. Ignoring air friction, the effect of
earth’s curvature, the physical size of the projectile and the width of the wall, find five appropriate launch angles so that the projectile passes through the opening.
Homework Equations
horizontal motion: x(t)=v*cos(theta)*t
vertical motion: y(t)=(v*sin(theta)*t)-(0.5*g*t^2)
time to land on ground: t= 2*v*sin(theta)/g
The Attempt at a Solution
i tried solving it by randomly taking bunch of theta angles and find horizontal distance and vertical distance. I believe the horizontal distance should be more than 40 and vertical distance should be between 30-40 m for a given theta angle so that the ball can go through opening. I am able to find various various of horizontal distance but i am getting very small values of vertical distance. I do not know what I am doing wrong.
Example: theta = 40
t=(2*35*sin(40))/9.8=4.6s
x(t)=35*cos(40)*4.6=123m
y(t)=(35*sin(40)*4.6)-(0.5*9.8*4.6^2)=-0.204?m...?this is wrong, I do not know why?
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