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Elina [12.6K]
3 years ago
9

1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.

Physics
1 answer:
Masja [62]3 years ago
4 0

Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

K=\frac{1}{2}mv^2

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

K=\frac{1}{2}(0.40)(16)^2=51.2  J

The potential energy of the ball is given by

U=mgh

where

m = 0.40 kg

g=9.8 m/s^2 is the acceleration of gravity

h = 22 m is the heigth of the cliff

Substituting,

U=(0.40)(9.8)(22)=86.2 J

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

E=K+U=51.2 + 86.2=137.4 J

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Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

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B=30°

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Answer:

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