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3 years ago

A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

2 answers:

7 0

IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy = (mass) x   (gravity) x (height)

   = (0.15 kg) x (9.8 m/s²) x (20 m)

   = 29.4 kg-m²/s² = 29.4 joules .

mel-nik [20]3 years ago

3 0

The conservation of energy says that energy in an isolated system remains constant, that it is not created nor destroyed but transforms from one form of energy to another.
So when the ball is thrown into the air, it has kinetic energy (KE). As it goes into the air, that kinetic energy transforms into gravitational potential energy (GPE). The higher it gets, the more GPE it has and therefore the less KE it has.
At the top, all of it is transformed into GPE and there is no KE.
But total energy at any point after the ball is thrown until it lands it has the same energy throughout.

Knowing this, you can set the beginning KE equal to the final GPE.
KE = GPE

Insert formula for GPE.

KE = mgh
KE = 0.15kg*9.8m/s^2*20.0m
KE = 29 J

The ball initially had 29 J of kinetic energy

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4 years ago

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If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the

GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = -10 m

$y_{0}$ : is the initial height = 0

$v_{0_{y}}$ : is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

$t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s$

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

$x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}$

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

$x_{f} = 5.51 m/s*1.43 s = 7.88 m$

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

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3 years ago

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Answer:

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Answer:

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3 years ago