Standing on the roof of a 62 m tall building, you throw a ball straight up with an initial speed of 14.5 m/s. If the ball misses the building on the way down, how long will it take from you threw the ball until it lands on the ground below? Give your answer in seconds and round the answer to three significant figures
To find the time it takes for the ball to land on the ground, we can use the equations of motion.
The key equation we can use is the equation for the vertical displacement of an object under constant acceleration:
Δy = v₀t + (1/2)at²
Where:
- Δy is the displacement (change in vertical position)
- v₀ is the initial velocity (14.5 m/s, upward)
- t is the time
- a is the acceleration due to gravity (-9.8 m/s², downward)
In our case, we want to find the time it takes for the ball to reach the ground, so the displacement (Δy) is equal to the height of the building (-62 m, negative because it's downward). We'll rearrange the equation to solve for time:
-62 = 14.5t + (1/2)(-9.8)t²
Let's solve this equation for t.
To find the time it takes for the ball to land on the ground below, we need to consider the motion of the ball in two parts: the time it takes for the ball to reach its maximum height while moving upward and the time it takes for the ball to fall from its maximum height to the ground.
First, let's determine the time it takes for the ball to reach its maximum height. We can use the equation of motion:
v = u + at
where:
- v is the final velocity (which is 0 m/s when the ball reaches its maximum height),
- u is the initial velocity (14.5 m/s directed upward),
- a is the acceleration due to gravity (-9.8 m/s², taking the downward direction),
- t is the time taken.
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Substituting the given values:
t = (0 - 14.5) / -9.8
t ≈ 1.4796 seconds
Now that we have the time it took for the ball to reach its maximum height, let's find the time it takes for the ball to fall from its maximum height to the ground. We can use the equation of motion:
s = ut + 0.5at²
where:
- s is the vertical distance traveled (62 m, equal to the height of the building),
- u is the initial velocity (0 m/s at the maximum height),
- a is the acceleration due to gravity (-9.8 m/s², taking the downward direction),
- t is the time taken.
Rearranging the equation, we get:
t = √(2s / a)
Substituting the given values:
t = √(2 * 62 / 9.8)
t ≈ 3.5596 seconds
Therefore, the total time it takes for the ball to land on the ground below is the sum of the time it took to reach the maximum height (1.4796 seconds) and the time it takes to fall from the maximum height (3.5596 seconds):
Total time ≈ 1.4796 + 3.5596
Total time ≈ 5.0392 seconds
Lastly, rounding the answer to three significant figures, the time it takes for the ball to land on the ground below is approximately 5.04 seconds.
V^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g = -(14.5^2)/-19.6 = 10.73 m.
Above the roof.
V = Vo + g*Tr = 0.
Tr = -Vo/g = -14.5/-9.8 = 1.48 s. = Rise
time.
0.5g*Tf^2 = 62+10.73.
4.9*Tf^2 = 72.73.
Tf^2 = 14.8.
Tf = 3.85 s. = Fall time.
Tr+Tf = 1.48 + 3.85 = 5.33 s. = Time in
flight.