Bull A is dropped form a balcony 19,6 m from the ground. At the same time an identical B is projected vertical downwards from the top of a building a 29, 6 m from the ground.

Calculating magnitude of the :
3.1 final velocity of the ball A
3.2 velocity with which B must be projected to reach the ground at a symptom as A

1 answer

To solve these problems, we can use the equations of motion for objects in free fall.

For both Bull A and Bull B, we will use the equation:
vf = vi + gt

where:
vf = final velocity
vi = initial velocity (in this case, for Bull A, vi = 0 and for Bull B, we need to find vi)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

To find the final velocity of Bull A, we can use the equation:
vf = vi + gt
Since Bull A is dropped from rest (vi = 0), we have:
vf = gt

Substituting the values, we get:
vf = (9.8 m/s^2)(19.6 m)
vf ≈ 192.08 m/s
Therefore, the final velocity of Bull A is approximately 192.08 m/s.

To find the velocity with which Bull B must be projected to reach the ground at the same time as Bull A, we need to find the time it takes for Bull A to reach the ground.

Using the equation:
vf = vi + gt
Since Bull A is dropped from rest (vi = 0) and Bull A reaches the ground (vf = 0), we have:
0 = (9.8 m/s^2)t
Solving for t, we get:
t = 0

This tells us that Bull A reaches the ground after 0 seconds.

Now, we can use the equation for Bull B:
vf = vi + gt

Substituting the values, we get:
0 = (9.8 m/s^2)t + vi

To have t = 0 (the time it takes for Bull A to reach the ground), we need the second term on the right side to be zero. Therefore, vi must also be 0.

This means that Bull B must be projected downwards from the top of the building with an initial velocity of 0 m/s to reach the ground at the same time as Bull A.