An object rolls off a tabletop with a horizontal velocity v0x = 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.
(b) How far, in meters, does the object land from the edge of the tabletop?
(c)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.
(d) What is the magnitude of the velocity (its speed) when it hits the floor?
(e) Find the angle of impact, in degrees below the horizontal.
(f) 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.
To solve these questions, we can use the kinematic equations of motion. The given information includes the initial horizontal velocity (v0x = 2.5 m/s), and the initial vertical height (y0 = 0.55 m) above the floor. We also know that the acceleration due to gravity is approximately constant at 9.8 m/s^2, and we can assume there is no air resistance.
(b) To find how far the object lands from the edge of the tabletop, we can use the equation for horizontal distance:
x = v0x * t
We need to find the time it takes for the object to hit the ground. To do this, we'll use the equation for vertical displacement:
y = y0 + v0y * t - (1/2) * g * t^2
Since the object is launched horizontally, the initial vertical velocity (v0y) is 0. Therefore, the equation simplifies to:
y = y0 - (1/2) * g * t^2
At the moment the object hits the ground, y = 0. We can substitute this information into the equation and solve for t:
0 = y0 - (1/2) * g * t^2
(1/2) * g * t^2 = y0
t^2 = (2 * y0) / g
t = √((2 * y0) / g)
Once we have the time, we can substitute it back into the equation for horizontal distance to find x.
(c) To find the vertical component of velocity when the object hits the ground, we can use the equation for vertical velocity:
v = v0y - g * t
Substituting the values we have, v0y = 0 and t = √((2 * y0) / g), we can calculate v.
(d) The magnitude of the velocity (speed) when the object hits the ground is simply the total velocity vector. It can be found using the Pythagorean theorem:
v = √(v0x^2 + v^2)
Substituting the values, we can calculate v.
(e) The angle of impact, θ, is the angle below the horizontal at which the object lands. This angle can be found using the equation:
tan(θ) = v / v0x
By taking the inverse tangent of both sides of the equation, we can solve for θ.
(f) To find the expression for the height of the object, y, in terms of x, v0x, y0, and g, we can use the equation for vertical displacement:
y = y0 + v0y * t - (1/2) * g * t^2
Since the object is launched horizontally, the initial vertical velocity (v0y) is 0. Therefore, the equation simplifies to:
y = y0 - (1/2) * g * t^2
To express y in terms of x, we need to eliminate t. We can use the equation for horizontal distance:
x = v0x * t
Rearranging this equation, we get:
t = x / v0x
Substituting this value of t into the equation for y, we can get the expression requested.