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:

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