A body is projected such that its kinetic energy at the top is 3/4 th of its initial kinetic energy.What is the initial angle of projection of the projectile with the horizontal ?
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As there is no acceleration in the horizontal direction the velocity component will remain same as vcos∅ and at highest vertical component will be zero. so total velocity at top will be vcos∅
K.E will be 1/2 m (vcos∅)²
initial energy will be 1/2 mv²
at the highest point
K.E = 3/4 of initial K.E
1/2 m(vcos∅)² = 1/2 mv²× 3/4
∅= 30°
K.E will be 1/2 m (vcos∅)²
initial energy will be 1/2 mv²
at the highest point
K.E = 3/4 of initial K.E
1/2 m(vcos∅)² = 1/2 mv²× 3/4
∅= 30°
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