Question

An object rolls off a tabletop with a horizontal velocity v0x 2 m s. The table is at a height yo 1.05 m. above the floor. Use a coordinate system with its origin on the floor directly beneath the point where the object rolls off the table, its horizontal x-axis lying directly beneath the object's trajectory, and its vertical y-axis pointing up(a) How long. in seconds. is the object falling before it hits the floor? (b) What's the horizontal distance, in meters, from the edge of the tabletop to where the object lands? (c) What is the vertical component of velocity in meters per second when the object hits the ground?

Answer 1

Answer:

A. t = 0.30s

B. s = 0.204m

C. Vfinal = 4.96 m/s

Explanation:

Using equations of motion,

Initial velocity, Vo = 2 m/s

Distance from the rable to the floor, S = 1.05 m

Acceleration due to gravity, a = 9.81m/s2

Using S = Vot + 1/2at^2

Inputting values,

1.05 = 2t + 1/2*(9.81)*t^2

4.9t^2 + 2t - 1.05 = 0

Solving the quadratic equation,

t = 0.30s

B. At the tabletop, Vo = Vfinal = 2m/s

Vo = 0m/s

Using the equation of motion,

Vfinal^2 = Vo^2 + 2as

Where s = horizontal distance

(2)^2 = 0^2 +2 * 9.81 * s

s = 4/19.62

= 0.204 m

C. For the vertical component, the initial velocity of the ball ia right at the maximum height before it increases vertically. Vfinal is the velocity when the object hust hit the ground.

Back to Vo = 2m/s

Vfinal = ?

S = 1.05 m

Using the equation of motion,

Vfinal^2 = Vo^2 + 2as

= 2^2 + 2*9.81*1.05

= 4 + 20.601

Vfinal = sqrt(24.601)

= 4.96m/s