Question

a ball whose kinetic energy is E is projected at angle of 45 degree to the horizontal what will be the kinetic energy of the ball at the highest point of its flight.

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Answer 1

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$<b>$
suppose the ball is projected with velocity u.Then,

$e = \frac{1}{2} m {v}^{2}$

At the highest point,the velocity of the ball will be
$\frac{1}{2} \times \frac{1}{2} m {u}^{2} = \frac{1}{2} e$

v = Horizontal component of the velocity of projection

= u cos45° = u√2

The k.e. of the ball at the highest point,

$e_{1} = \frac{1}{2}m {v}^{2} = \frac{1}{2 } m( \frac{ {u}}{ \sqrt{2} } )^{2}$

$\frac{1}{2} \times \frac{1}{2} m {u}^{2} = \frac{1}{2} e$
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Answer 2

hey there♠♠♠♠


➖➖➖here's ur answer➗➗➗

at the angle, tita=0°,, the velocity component with horizontal is given by,

u cos 0° = u (1)= u

SIMILARLY , the KINETIC energy was given by

K= 1/2 m (u cos 0°)^2

K = 1/2 m u^2====>>(1)

then, when the horizontal component of VELOCITY was elevated with an angle of 45°,, then its VELOCITY component is given by,,

u cos45°

then ucos45 = u × 1/√2


then kinetic energy is given by,,,,

K' = 1/2 m (u cos45)^2

= 1/2 m u^2(1/2)

= 1/4 m u^2===>>(2)

from equation 1 and 2


which implies 2K' = K or K' =1/2 K, that is Kinetic energy at an angle 0° is two times of Kinetic energy at 45°,,,

==extra===>
which implies that KINETIC energy decreases with height.....