Levon throws a ball straight upwards with an initial velocity of 15.0 m/s.
a.How high will it reach?
b.How long will it take unit his friend in a tree 5.0 m above him catches the ball? (he catches the ball on its way back down)
a. V^2 = Vo^2 + 2g*h = 0
h = -Vo^2/2g = -225/-19.6 = 11.5 m.
b. V = Vo + g*Tr = 0.
Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise
time.
0.5g*Tf^2 = 11.5-5 = 6.5 m.
4.9Tf^2 = 6.5
Tf^2 = 1.33
Tf = 1.15 s. = Fall time.
Tr+Tf = 1.53 + 1.15 = 2.68 s. To catch
the ball.
To answer these questions, we can use the equations of motion for an object in free fall (assuming no air resistance).
a. To find the maximum height, we need to determine the time it takes for the ball to reach its peak and then use that time to find the height.
First, let's find the time it takes for the ball to reach its peak. We can use the equation:
v = u + at
where:
v = final velocity (at the peak, the velocity is 0)
u = initial velocity (15.0 m/s)
a = acceleration due to gravity (approximately -9.8 m/s²)
Rearranging the equation, we get:
t = (v - u) / a
Plugging in the values, we have:
t = (0 - 15.0) / -9.8
t ≈ 1.53 seconds
Now that we have the time, we can find the height using the equation:
s = ut + (1/2)at²
where:
s = displacement (maximum height)
t = time (1.53 seconds)
u = initial velocity (15.0 m/s)
a = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we have:
s = 15.0 * 1.53 + (1/2) * -9.8 * (1.53)²
s ≈ 11.29 meters
Therefore, the ball will reach a height of approximately 11.29 meters.
b. To determine the time until Levon's friend catches the ball, we need to find the time it takes for the ball to reach a height of 5.0 meters above Levon's initial position.
Using the same equation as before:
s = ut + (1/2)at²
where:
s = displacement (5.0 meters)
u = initial velocity (15.0 m/s)
a = acceleration due to gravity (-9.8 m/s²)
Rearranging the equation, we get:
t² - (2u/a)t + 2s/a = 0
This is a quadratic equation in terms of t. Plugging in the values, we have:
t² - (2 * 15.0 / -9.8) * t + (2 * 5.0 / -9.8) = 0
Now we can solve this quadratic equation to find the roots (t values). The positive root will give us the time it takes for the ball to reach 5.0 meters above Levon's initial position.
Using the quadratic formula, we have:
t = (-b ± sqrt(b² - 4ac)) / 2a
Plugging in the values, we get:
t = [-(2 * 15.0 / -9.8) ± sqrt((2 * 15.0 / -9.8)² - 4 * 1 * (2 * 5.0 / -9.8))] / 2 * 1
Simplifying the equation, we have:
t ≈ [3.06 ± sqrt(3.06² + 0.408)] / 2
Solving for t, we have two solutions:
t ≈ (3.06 + sqrt(3.06² + 0.408)) / 2 ≈ 2.19 seconds
t ≈ (3.06 - sqrt(3.06² + 0.408)) / 2 ≈ 0.64 seconds
Since Levon's friend catches the ball on its way back down, the time it takes until his friend catches the ball is approximately 0.64 seconds.
Therefore, his friend will catch the ball after approximately 0.64 seconds.