A ball is thrown down from the top of a building with a speed of 19.6m/s. The top of the building is 39.2m above the level of the sidewalk. The ball needs 1.46s to travel from the top of the building to the sidewalk. What is the velocity of the ball just before the ball reaches the sidewalk?
final KEnergy =initial PEnergy+initialKEnergy
1/2 m vf^2 = mg*39.2 + 1/2 m 19.6^2
solve for vf
To find the velocity of the ball just before it reaches the sidewalk, we can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the ball is thrown downward, so the initial velocity is -19.6 m/s (negative because it is in the opposite direction of the positive direction). The acceleration due to gravity is -9.8 m/s² (negative because it acts downward). The time taken to travel from the top of the building to the sidewalk is 1.46 seconds.
Substituting these values into the equation, we have:
v = -19.6 m/s + (-9.8 m/s²)(1.46 s)
By simplifying,
v = -19.6 m/s - 14.308 m/s
v = -33.908 m/s
Therefore, the velocity of the ball just before it reaches the sidewalk is -33.908 m/s. The negative sign indicates that the velocity is directed downward.