Question 222: 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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