A body is projected at an angle of projection Θ with kinetic energy E. Neglecting air friction, the kinetic energy at the highest point is a) zero (b) E (c) E cos Θ (d) E cos^2 Θ please tell the answer and explain it..
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When the body reaches the highest point it has 0 velocity in the vertical but will still have a component of velocity in the horizontal (for other than theta of 90 degrees).
E = (1/2)m*v^2 ... the standard equation for kinetic energy
v*cos(theta) = the component of velocity in the horizontal
The horizontal velocity is constant since there is no air resistance, etc so at max altitude:
E = (1/2)m*[v(cos(theta)]^2 = [(1/2)*m*v^2]*cos^2(theta) = E*cos^2(theta)
So the answer is d
E = (1/2)m*v^2 ... the standard equation for kinetic energy
v*cos(theta) = the component of velocity in the horizontal
The horizontal velocity is constant since there is no air resistance, etc so at max altitude:
E = (1/2)m*[v(cos(theta)]^2 = [(1/2)*m*v^2]*cos^2(theta) = E*cos^2(theta)
So the answer is d