Answer:

A ball is rolled off the edge of a dining table 1.5 m high with a horizontal velocity of 2.5 m/s . With what velocity does it strike the ground? How far from the edge of the table did it land on the floor?

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Eduardo Dequilla

Answered March 4, 2021

There are three succeeding steps and three formulas to use in this particular problem. First, there is a need to solve for the time the ball has fallen to the from the table top. Next. We use the velocity formula using the time the ball has fallen and third we need to use the dx formula to calculate for the horizontal distance of the ball from the edge of the table.

Solving for the time the ball has fallen

t = sqrt ((2 * height) / g)

t = sqrt((2 * 1.5 m) / 9.8 m/s^2)

t = sqrt ( 3.0 m / 9.8 m/s^2)

t = sqrt ( 0.3061)

t = 0.553 seconds

Solving for the horizontal and vertical velocities velocity

For the horizontal velocity it is already given. Vx = 2.5 m/s.

Solving for the final vertical velocity

Vfy = g * t where Vfy is the final vertical velocity, g = 9.8 m.s^2 and t = 0.553 seconds.

Vfy = 9.8 m.s^2 * 0.553 s = 5.42 m/s

Solving for the combined velocities V

V^2 = Vx^2 + Vfy^2

V^2 = (2.5 m/s)^2 + (5.42)^2

V^2 = 35.63 (m/s)^2

V = sqrt (35.63)

V = 5.97 m/s

Calculating the horizontal distance of the ball from the edge of the table

dx = Vx * t

dx = 2.5 m/s * 0.553 s

dx = 1.38 m

With what velocity does it strike the ground?

The velocity of the ball upon striking the ground is 5.97 m/s.

How far from the edge of the table did it land on the floor?

The ball landed on the floor at a horizontal distance of 1.38 m from the edge of the table.

Explanation:

hope it helps ^-^