A ball is thrown upward from the top of a 50.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 44.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?
Tr = (Vf-Vo)/g,
Tr = (0-12) / -9.8 = 1.22 s. = Time to
rise to max ht.
hmax = ho + Vo*t + 0.5g*t^2,
hmax=50 + 12*1.22 - 4.9(1.22)^2=57.35 m
= ht. above gnd.
hmax = Vo*t + 0.5g*t^2 = 57.35 m.
0 + 4.9t^2 = 57.35,
t^2 = 11.70,
Tf = 3.42 s. = Time to fall to gnd.
d = V*Tf,
V = d/Tf = 44 / 3.42 = 12.9 m/s.=Speed
of the person.
To find the average speed of the person, we need to determine the time it takes for the ball to fall from the top of the building to the bottom.
1. Let's first find the time it takes for the ball to reach its maximum height. We can use the formula for the time of flight for a vertically thrown object:
- Initial velocity (u) = 12.0 m/s
- Final velocity (v) = 0 m/s (at the highest point)
- Acceleration (a) = -9.8 m/s^2 (negative because it opposes the direction of motion)
- Using the formula v = u + at, we can solve for time (t) when the ball reaches its maximum height.
v = u + at
0 = 12 + (-9.8)t
-12 = -9.8t
t = -12 / -9.8
t ≈ 1.22 seconds
2. Now, let's find the total time for the ball to fall from the top of the building to the bottom. Since the time it takes to reach the maximum height is the same as the time to fall back down, the total time is twice the time to reach the maximum height.
Total time = 2 * t ≈ 2 * 1.22 ≈ 2.44 seconds
3. The person needs to reach the bottom of the building in the same time as the ball falls. The distance the person needs to cover is 44.0 m.
Average speed = Total distance / Total time
Average speed = 44.0 m / 2.44 s
Average speed ≈ 18.03 m/s
Therefore, the average speed of the person must be approximately 18.03 m/s to catch the ball at the bottom of the building.
To find the average speed of the person, we need to determine the time it takes for the ball to fall from the top of the building to the ground. Since the ball is thrown vertically upwards, it will eventually reach its maximum height and then fall back down.
First, let's calculate the time it takes for the ball to reach its maximum height. We can use the equation:
t = Vf / g,
where t is the time, Vf is the final velocity (which is 0 when the ball reaches its maximum height), and g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
Using the equation, we can find:
t = 12.0 m/s / 9.8 m/s^2 = 1.22 seconds.
Next, let's calculate the total time for the ball to reach the ground. Since the ball takes the same amount of time to reach its maximum height and fall back down, the total time is twice the time calculated above:
Total time = 2 * 1.22 seconds = 2.44 seconds.
Now, let's calculate the average speed of the person. The person needs to cover a distance of 44.0 m in 2.44 seconds to catch the ball.
Average speed = Total distance / Total time.
Average speed = 44.0 m / 2.44 s = 18.03 m/s.
Therefore, the average speed of the person must be approximately 18.03 m/s to catch the ball at the bottom of the building.