Object Rolling off a table top

Question

An object rolls off a tabletop with a horizontal velocity v_0x = 2.5 m/s. The table is at a height y₀ = 0.55 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.

How long, in seconds, is the object falling before it hits the floor?

How far, in meters, does the object land from the edge of the tabletop?

What is the vertical component of velocity, in meters per second, when the object hits the ground? Recall that the positive y-direction is upwards.

What is the magnitude of the velocity (its speed) when it hits the floor?

Find the angle of impact, in degrees below the horizontal.

Enter an expression for the height of the object y, in terms of x, v0x, y0, and g. This expression should not depend on the time.

Answer 1

Given:

V_ox = 2.5 m/s

V_oy = 0 m/s

Y_o = 0.55 m

Xo = 0 m

Y = 0 m

g = - 9.81 m/s^2

Solution

a) How long until it hits the floor?

Y = -(1/2) * g * t^2 + V_oy t + Y_o

_{Y = 0 m}

V_oy = 0 m/s

This will simplify the equation

0 = -(1/2) * g * t^2 +Y_o

t^2 = (2 * Yo) / g

Plug in givens and solve for time, t = 0.334 seconds

b) How far in meters does the object land from the edge of the table top?

This is the horizontal trajectory, we apply the kinematic equation for horizontal motion.

X = Vox * t + Xo

Apply givens

X = Vox * t

Solve for the horizontal distance...

X = 2.5 * 0.334 = 0.835 m

c) Vertical component of velocity as it hits the floor

2a (y-yo) + Voy ^2 = Vy^2

Apply givens

-2g(-yo) = Vy^2

2* g * Yo = Vy^2

2 * 9.81 * 0.55 = Vy^2

Vy = 3.285 m/s (Downward direction)

d) Magnitude of velocity when it hits the floor...

Vox = Vx = Constant = 2.5 m/s (Horizontal velocity component is constant throughout motion)

Vy = 3.285 m/s

Solve for magnitude of vector

V = (Vox^2 + Vy^2)^1/2

V = 4.128 m/s

e) Find the angle of impact

Draw a vector diagram

Velocity on the X points to the right

Velocity on the Y points downward

Magnitude of Velocity is on the fourth quadrant

tan (theta) = Vy/Vox

Theta = tan-1(Vy/Vox) = -52.72 degree (Measured from positive X-axis, clockwise)

f) Write an expression for height of object with respect to variables mentioned in problem statement

Start with the following equation

X = Vox * t

Solve for variable t

t = X/Vox

Now using this variable plug into

Y = -(1/2) * g * t^2 + Yo

Please note Voy is crossed off from the equation, since it is zero

Y = -(1/2)*g*(X/Vox)^2 + Yo

This equation meets the criteria.It is the height as a function of the variables requested....