A ball is thrown directly upward with an initial velocity of 15 m/s from the rooftop of a building that is 30 meters in height. It is caught by a student whose hand is extended from a window. If the students hand is located 18 meters below the rooftop. What is the ball’s velocity at t=3.5 seconds in m/s
To solve this problem, we can use the equations of motion for uniformly accelerated linear motion.
The initial velocity of the ball is 15 m/s, and it is thrown directly upward. Its acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).
First, we can calculate the time it takes for the ball to reach its maximum height. The equation we can use is:
v = u + at
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (15 m/s)
a = acceleration (-9.8 m/s^2)
t = time
0 = 15 + (-9.8)t
-15 = -9.8t
t = -15 / -9.8
t ≈ 1.53 s
So, it takes approximately 1.53 seconds for the ball to reach its maximum height. Now, we can calculate the ball's velocity at t = 3.5 seconds.
First, we find the time the ball takes to fall from its maximum height to the student's hand. The height is equal to the height of the building minus the distance of the student's hand from the rooftop.
height = building height - distance of hand from rooftop
height = 30 - 18
height = 12 meters
Using the equation:
height = ut + (1/2)at^2
Where:
u = initial velocity (0 m/s at maximum height)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
12 = 0 + (1/2)(-9.8)t^2
24 = -9.8t^2
t^2 = 24 / -9.8
t^2 ≈ -2.45
t ≈ √(-2.45) [Ignoring the negative square root, since time cannot be negative]
t ≈ √2.45
t ≈ 1.57 seconds
Therefore, the ball takes approximately 1.57 seconds to fall from its maximum height to the student's hand. Adding this time to the time it took to reach the maximum height (1.53 seconds), we get the total time:
Total time = time to reach the maximum height + time to fall from maximum height to the student's hand
Total time ≈ 1.53 + 1.57
Total time ≈ 3.1 seconds
Now, we can calculate the ball's velocity at t = 3.5 seconds using the equation:
v = u + at
where:
v = final velocity
u = initial velocity (15 m/s)
a = acceleration (-9.8 m/s^2)
t = time
v = 15 + (-9.8)(3.5 - 3.1)
v = 15 + (-9.8)(0.4)
v = 15 + (-3.92)
v ≈ 11.08 m/s
Therefore, at t = 3.5 seconds, the ball's velocity is approximately 11.08 m/s.