Asked by blue ivy
A 0.250 kg ball is thrown straight upward with an initial velocity of 38 m/s. If air friction is ignored, calculate the:
(a) height of the ball when its speed is 12 m/s
(b) height to which the ball rises before falling
(c) How would your answers to (a) and (b) change if
you repeated the exercise with a ball twice as
massive?
(a) height of the ball when its speed is 12 m/s
(b) height to which the ball rises before falling
(c) How would your answers to (a) and (b) change if
you repeated the exercise with a ball twice as
massive?
Answers
Answered by
bobpursley
Use energy considerations:
a) 1/2 m vi^2=1/2 mg vf^2+mg*hfinal
b) same equation, vf=0
c) divide the equation in a) by m, and see how the mass then affects that equation.
a) 1/2 m vi^2=1/2 mg vf^2+mg*hfinal
b) same equation, vf=0
c) divide the equation in a) by m, and see how the mass then affects that equation.
Answered by
angel
A ball is thrown straight up from ground level. After a time 4.1 s, it passes a height of 157.1 m. What was its initial speed? The accelera-
tion due to gravity is 9.8 m/s2 . what is its initial speed
tion due to gravity is 9.8 m/s2 . what is its initial speed
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