Levon throws a ball straight upwards with an initial velocity of 15.0 m/s.a) How high will it reach? (Answer: 11.5 m)b) How long will it take until his friend in a tree 5.0 m above him catches the ball? (Hecatches the ball on its way back down.)
Question
Levon throws a ball straight upwards with an initial velocity of 15.0 m/s.a) How high will it reach? (Answer: 11.5 m)b) How long will it take until his friend in a tree 5.0 m above him catches the ball? (Hecatches the ball on its way back down.)
Solution
a) To find out how high the ball will reach, we can use the following formula from physics:
h = (v^2) / (2g)
where:
- h is the height,
- v is the initial velocity, and
- g is the acceleration due to gravity.
Given that v = 15.0 m/s and g = 9.8 m/s^2, we can substitute these values into the formula:
h = (15.0 m/s)^2 / (2 * 9.8 m/s^2) = 11.5 m
So, the ball will reach a height of 11.5 m.
b) To find out how long it will take until his friend in a tree 5.0 m above him catches the ball, we need to calculate two times: the time it takes for the ball to reach its maximum height, and the time it takes for the ball to fall from its maximum height to the height of the tree.
- The time it takes for the ball to reach its maximum height can be calculated using the formula:
t = v / g
Substituting the given values:
t = 15.0 m/s / 9.8 m/s^2 = 1.53 s
- The time it takes for the ball to fall from its maximum height to the height of the tree can be calculated using the formula:
t = sqrt((2h) / g)
where h is the difference between the maximum height and the height of the tree (11.5 m - 5.0 m = 6.5 m). Substituting the values:
t = sqrt((2 * 6.5 m) / 9.8 m/s^2) = 1.15 s
Adding these two times together gives the total time:
1.53 s + 1.15 s = 2.68 s
So, it will take approximately 2.68 seconds until his friend in the tree catches the ball.
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