Asked by joleen
A woman throws a cricket ball upwards into the air. The ball leaves the womans hand at 1.8m above the ground at speed of 9.7. It rises and then falls back to the ground. Considering the energies involved calculate the speed at the point it hits the ground.
accerlation due to gravity (g) is 9.8 m s, neglect air resistance
accerlation due to gravity (g) is 9.8 m s, neglect air resistance
Answers
Answered by
MathMate
Equate final kinetic energy to initial kinetic and potential energies.
Initial:
KE=(1/2)m(9.7)²
PE=mg(1.8)
Final:
KE=(1/2)mv²
PE=0
All energies are in joules if you work with metres, kilograms and seconds.
Initial:
KE=(1/2)m(9.7)²
PE=mg(1.8)
Final:
KE=(1/2)mv²
PE=0
All energies are in joules if you work with metres, kilograms and seconds.
Answered by
joleen
Is that the same as
Ek = ½mv² - ½mu²
or does mgh have to be in the equation?
Ek = ½mv² - ½mu²
or does mgh have to be in the equation?
Answered by
MathMate
mgh has to be in the equation because the ball was initially 1.8m higher than the ground. The velocity required is at ground level.
(1/2)mv²=(1/2)mu²+mgh
u,v being initial and final velocities.
(1/2)mv²=(1/2)mu²+mgh
u,v being initial and final velocities.
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