A person standing on a building ledge throws a ball vertically from a launch position 55 m above the ground. It takes 2.0 s for the ball to hit the ground. With what initial speed was the ball thrown?

The initial speed at which the ball is thrown upward is 17.7 m/s.

According to the question the initial position of the ball, y = 55 m, and the final position of the ball is y' = 0 m. We have assumed upward direction as positive direction and downward as negative direction.

so the total displacement:

d = y'-y = 0 - 55

d = -55 m

now applying the second equation of motion:

d = ut - (1/2)gt²

where t = 2s ( given ) and g = 9.8 m/s².

-55 = 2u - 0.5×9.8×4

-55 = 2u - 19.6

u = -17.7 m/s

the negative sign indicated that the initial velocity is opposite to the direction of displacement.

This means the initial velocity is upward as it should be.

Therefore, the initial speed of the ball is 17.7m/s

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Vo = 17.69 [m/s]

Explanation:

To solve this problem we must use two equations of kinematics.

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 55 [m]

t = time = 2 [s]

Now we replace the gravity acceleration into the second equation:

And then into the first equation:

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.