JEE MAIN - Physics (2002 - No. 54)
A ball whose kinetic energy E, is projected at an angle of $$45^\circ $$ to the horizontal. The kinetic energy of the ball at the highest point of its height will be
E
$${E \over {\sqrt 2 }}$$
$${E \over 2}$$
zero
Explanation
Assume the ball of mass m is projected with a speed u. Then the kinetic energy(E) at the point of projection = $${1 \over 2}m{u^2}$$
At highest point of flight only horizontal component of velocity $$u\cos \theta $$ present as at highest point vertical component of velocity is = 0.
Note : The horizontal component of velocity does not change in entire projectile motion.
At highest point the velocity is = $$u\cos \theta $$ = $$u\cos 45^\circ $$ = $${u \over {\sqrt 2 }}$$
$$\therefore$$ The kinetic energy at the height point = $${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$$
= $${1 \over 2}m{u^2} \times {1 \over 2}$$ = $${E \over 2}$$
At highest point of flight only horizontal component of velocity $$u\cos \theta $$ present as at highest point vertical component of velocity is = 0.
Note : The horizontal component of velocity does not change in entire projectile motion.
At highest point the velocity is = $$u\cos \theta $$ = $$u\cos 45^\circ $$ = $${u \over {\sqrt 2 }}$$
$$\therefore$$ The kinetic energy at the height point = $${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$$
= $${1 \over 2}m{u^2} \times {1 \over 2}$$ = $${E \over 2}$$
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