Question

a cricket ball is hit at 45 degree to the horizontal with a kinetic energy E. calculate the kinetic energy at the highest point

Answer 1

Answer:

The kinetic energy at the highest point is E/2

Explanation:

Let the mass of the ball be m

When the ball is hit, its motion will be projectile motion. If its initial velocity was u then the initial kinetic energy of the ball = $\frac{1}{2}mu^{2}$

Thus, $E=\frac{1}{2} mu^{2}$

Since, in projectile motion the acceleration is only due to gravity in the vertical direction and in the horizontal direction there is no acceleration.

Therefore the horizontal component of the velocity does not change. At the highest point, the vertical component of the velocity will be zero and only the horizontal component will remain which will be equal to the initial horizontal component of the velocity as there is no acceleration in the horizontal direction

Horizontal component of the velocity = ucos45° = $\frac{u}{\sqrt{2} }$

Therefore, the kinetic energy at the highest point

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

$=\frac{1}{2} (\frac{1}{2}mu^{2})$

$=\frac{E}{2}$

Answer 2

Answer:

Kinetic Energy at Highest point = E/2

Explanation:

a cricket ball is hit at 45 degree to the horizontal with a kinetic energy E. calculate the kinetic energy at the highest point

Let say Initial Velocity = V

mass of Ball = m

Kinetic Energy = (1/2) m V² = E       Eq1

Horizontal Velocity = VCosα = VCos45° = V/√2

Vertical Velocity = VSinα = VSin45° = V/√2

at Highest Point Vertical Velocity becomes 0

& Horizontal Velocity remains same V/√2

Kinetic Energy at Highest point = (1/2)m (V/√2)²

=  (1/2)mV²/2

from eq 1 :  (1/2) m V²  = E

= E/2

Kinetic Energy at Highest point = E/2