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A ball is thrown upward with a speed of 40 m/s. Approximately how much time goes it take the ball to travel from the release location
Question
A ball is thrown upward with a speed of 40 m/s. Approximately how much time goes it take the ball to travel from the release location (A) to its highest point(B)? Approximately how much total time is the ball in the air before it returns to back to its original height (C)?
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Physics
3 years
2021-07-17T12:23:15+00:00
2021-07-17T12:23:15+00:00 1 Answers
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Answers ( )
Answer:
(I). The time at highest point 4.0 sec.
(II). It returns to back to its original height in 8.1 sec
Explanation:
Given that,
Velocity![Rendered by QuickLaTeX.com v=40\ m/s](https://documen.tv/wp-content/ql-cache/quicklatex.com-2a11d7c54b94652783263885c4ddea72_l3.png)
(I). We need to calculate the time at highest point
Using equation of motion
Where, v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time
Put the value into the formula
(II). We know that, when the ball to travel from the initial point and reached at initial point then the displacement is zero.
We need to calculate the total time when it returns to back to its original height
Using equation of motion
Where, s = displacement
g = acceleration due to gravity
t = time
u = velocity
Put the value in the equation
Hence. (I). The time at highest point 4.0 sec.
(II). It returns to back to its original height in 8.1 sec