You shoot a cannon with a vi of 18 m/s. You need to get it 16m and over a 10m wall. The acceleration in the x is 0 and in the y it is gravity. The question is at what angle theta can you shoot the cannon over the fence and 16 meters away. How do I find theta?
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cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
The length of the arc is r*theta where r is the radius and theta the angle subtended by the arc at the centre of the circle. If you do not know theta (or cannot derive it), you cannot find the length of the arc.
sine[theta]=opposite/hypotenuse=square root of (1-[cos[theta]]^2)
tan(theta) = 1 then theta = tan-1(1) + n*pi where n is an integer = pi/4 + n*pi or pi*(1/4 + n) Within the given range, this gives theta = pi/4 and 5*pi/4
The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.