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Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part

earnstyle [38]

The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

a) The kinetic energy of the ball immediately after it is hit

b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

Answer: a) KE = 131.64 J

b) KE = 0

c) h = 126 ft

Explanation: Kinetic energy is the energy an object posses due to its motion. It can be calculated as $KE=\frac{1}{2}mv^{2}$

a) Kinetic energy's unit is Joule. So, we have to transform ounce in kg and ft/s in m/s for the units to correspond:

m = 5.1(0.02835)

m = 0.1445 kg

v = 140 ft = 42.67 m/s

Then, kinetic energy is

$KE=\frac{1}{2}(0.1445)(42.67)^{2}$

KE = 131.64 J

Kinetic energy immediately after the ball is hit is 131.64 J.

b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

c) This type of motion is projectile motion. The maximum height on a projectile motion can be determined by

$v_{y}^{2}=v_{0y}^{2}-2g\Delta y$

When h is maximum, $v_{y}=0$

Velocity of the ball has an angle with the horizontal, so initial velocity at the y-axis is

$v_{0y}=v_{0}sin(\theta)$

Substituting and solving

$v_{y}^{2}=v_{0}^{2}sin^{2}(\theta)-2gh$

$0=(42.67)^{2}sin^{2}(40)-2(9.8)h$

$19.6h=(42.67)^{2}(0.643)^{2}$

$h=\frac{(1820.73)(0.4132)}{19.6}$

h = 38.4 m

Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

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