Asked by joleen
If you throw a cricket ball up into the air and the ball leaves your hand at 1.8m and travel with a speed of 9.7m s. What is the speed at the point it hits the floor. Accerlation due to gravity is 9.8 ms no air resistance
Answers
Answered by
drwls
No matter what angle you throw it at, the kinetic energy increases by
M g *1.8 m when it hits the ground. That means
(M/2)(V2^2 - V1^2) = M*g*1.8
Note that the mass M cancels.
V2^2 = 2*g*1.8m + V1^2
= 37.04 m^2/s^2 + V1^2
= 37.04 + 94.1
Solve for V1
M g *1.8 m when it hits the ground. That means
(M/2)(V2^2 - V1^2) = M*g*1.8
Note that the mass M cancels.
V2^2 = 2*g*1.8m + V1^2
= 37.04 m^2/s^2 + V1^2
= 37.04 + 94.1
Solve for V1
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