Question 212: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

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 $$u\cos 	heta $$ 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 = $$u\cos 	heta $$ = $$u\cos 45^\circ $$ = $${u \over {\sqrt 2 }}$$
$$	herefore$$ The kinetic energy at the height point = $${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} ight)^2}$$
= $${1 \over 2}m{u^2} 	imes {1 \over 2}$$ = $${E \over 2}$$

Answer

E

Answer

$${E \over {\sqrt 2 }}$$

Answer

$${E \over 2}$$

Answer

zero

Work, Energy and Power & Circular Motion

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

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