Golf Ball Released from Tall Building: Velocity & Impact

[prowlerplayer]

A golf ball is released from rest from the top
of a very tall building. Choose a coordinate
system whose origin is at the starting point
of the ball, and whose y-axis points vertically
upward.
The acceleration of gravity is 9.8 m/s
2
.
Neglecting air resistance, calculate the ve-
locity of the ball after 3.99 s. Answer in units
of m/s. 029 (part 2 of 4) 10.0 points
What is its speed just before impact with the
ground? Answer in units of m/s.
030 (part 3 of 4) 10.0 points
Now assume the hot air balloon is traveling
vertically downward at a constant speed of
4.9 m/s.

 

[jambaugh]

Please submit homework in the proper format. See:
https://www.physicsforums.com/showpost.php?p=1042781&postcount=2"

Without some attempt at as solution you won't get any help here.

 

[tomcue]

What is the speed of the ball just
after it is released from the balloon

I would approach this scenario by first setting up a coordinate system with the origin at the starting point of the ball and the y-axis pointing vertically upward. This will allow us to easily track the motion of the ball as it falls.

Next, we can use the equation for position in terms of time and acceleration to calculate the velocity of the ball after 3.99 seconds. Since the ball is released from rest, its initial velocity is 0 m/s. Using the equation x = x0 + v0t + 1/2at^2, we can plug in the values to get x = 0 + 0 + 1/2(9.8)(3.99)^2 = 77.9 m. This is the position of the ball after 3.99 seconds, but we are interested in its velocity. So, we can use the equation v = v0 + at to find the velocity at this position. v = 0 + (9.8)(3.99) = 39.0 m/s. This is the velocity of the ball after 3.99 seconds.

Next, we can calculate the speed of the ball just before impact with the ground. Since the ball is falling vertically, its final velocity will only have a vertical component. We can use the equation v^2 = v0^2 + 2ad to find the final velocity. Since the ball is starting from rest, v0 = 0. Plugging in the values, we get v^2 = 0 + 2(9.8)(77.9) = 1523.24. Taking the square root, we get v = 39.0 m/s. This is the speed of the ball just before impact with the ground.

Finally, if we assume that the hot air balloon is traveling vertically downward at a constant speed of 4.9 m/s, we can use the same equation v^2 = v0^2 + 2ad to find the speed of the ball just after it is released from the balloon. Since the ball is starting from rest again, v0 = 0. Plugging in the values, we get v^2 = 0 + 2(9.8)(4.9) = 96.04. Taking the square root, we get v =