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3 years ago

What happens to its kinetic energy when its reaches the maximum height ?

1 answer:

7 0

when it reaches the maximum height, all the energy has now been converted into potential energy.when a ball is thrown straight upto into the air,all its initial kinetic energy converted into gravitational potential energy when it reaches its maximum height

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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.

6 0

3 years ago

You place a small air cart on a 4.00 m air track and set it into motion with a speed of 0.75 m/s. Due to very good elastic bumpe

san4es73 [151]

Answer:

6 seconds

0.167 Hz

Explanation:

Distance = 4 m

Velocity = 0.75 m/s

Period is given by

$T=2\times Distance\times Velocity\\\Rightarrow T=2\times 4\times 0.75\\\Rightarrow T=6\ s$

The period of the motion is 3 seconds

Frequency is given by

$f=\frac{1}{T}\\\Rightarrow f=\frac{1}{6}\\\Rightarrow f=0.167\ Hz$

The frequency of the motion is 0.33 Hz

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svetoff [14.1K]

I believe the answer would be elastic because you are pulling the sting of the bow back.

hope this helps :)

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4 years ago

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3 years ago

What happens to its kinetic energy when its reaches the maximum height ?​

What happens to its kinetic energy when its reaches the maximum height ?