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.
(b) The horizontal distance traveled by the object can be found using the formula for horizontal distance:
x = v0x * t
Since the object falls vertically, the time taken to hit the ground can be found using the formula for vertical displacement:
y = y0 + v0y * t - (1/2) * g * t^2
Where v0y is the initial vertical component of velocity and g is the acceleration due to gravity.
To find t, we need to find v0y first. Since the object rolls off the table with no initial vertical velocity, v0y = 0.
Therefore, the equation for y becomes:
y = y0 - (1/2) * g * t^2
When the object hits the ground, y = 0. Solving for t gives:
t = sqrt(2 * y0 / g)
Now we can calculate x using the value of t:
x = v0x * t
Plugging in the given values:
x = 2.5 m/s * sqrt(2 * 0.55 m / 9.8 m/s^2)
x ≈ 0.936 m
Therefore, the object lands approximately 0.936 meters from the edge of the tabletop.
(c) The vertical component of velocity when the object hits the ground can be found using the equation:
v = v0y - g * t
Since v0y = 0 and we have already calculated the value of t, we can substitute and solve:
v = - g * t
v = - 9.8 m/s^2 * sqrt(2 * 0.55 m / 9.8 m/s^2)
v ≈ - 1.171 m/s
The negative sign indicates that the vertical component of velocity is in the downward direction.
(d) The magnitude of velocity (speed) when the object hits the floor can be found using the formula for speed:
speed = sqrt(vx^2 + vy^2)
Since the horizontal component of velocity remains constant throughout the motion, vx = v0x. And we have already calculated vy:
speed = sqrt((2.5 m/s)^2 + (-1.171 m/s)^2)
speed ≈ 2.761 m/s
Therefore, the magnitude of the velocity when it hits the floor is approximately 2.761 meters per second.
(e) The angle of impact, in degrees below the horizontal, can be found using the equation:
tan(Θ) = vy / vx
Plugging in the calculated values:
tan(Θ) = (-1.171 m/s) / (2.5 m/s)
Θ ≈ -25.12°
Therefore, the angle of impact is approximately 25.12 degrees below the horizontal.
(f) The expression for the height of the object y, in terms of x, v0x, y0, and g, can be derived from the equation for vertical displacement:
y = y0 + v0y * t - (1/2) * g * t^2
Since v0y = 0 (the object rolls off the table with no initial vertical velocity), the equation simplifies to:
y = y0 - (1/2) * g * t^2
We have already calculated the value of t:
y = y0 - (1/2) * g * ((2 * y0 / g))
y = y0 - y0
y = 0
Therefore, the expression for the height of the object y is y = 0, which means that at any horizontal distance x from the edge of the tabletop, the object will always be at a height of 0 (on the floor).
To solve these questions, we can use the kinematic equations of motion:
(a) The horizontal distance traveled by the object can be found using the equation:
x = v0x * t
Since the object rolls off the table horizontally, there is no initial vertical velocity (v0y = 0) and no vertical acceleration (ay = 0). Therefore, we can ignore the vertical motion entirely and focus only on the horizontal motion.
(b) The time it takes for the object to reach the ground can be found using the equation:
y = y0 + v0y * t + (1/2) * ay * t^2
Since the table height is y0 = 0.55 m and there is no initial vertical velocity or acceleration, the equation simplifies to:
y = y0 = 0 + 0 + 0
Solving for t:
0 = 0.55 + 0.5 * (-9.81) * t^2
-0.5 * 9.81 * t^2 = 0.55
t^2 = 0.55 / (0.5 * 9.81)
t^2 = 0.112
t ≈ √(0.112)
t ≈ 0.335 s
(c) To find the vertical component of velocity when the object hits the ground, we can use:
v = v0y + ay * t
Since v0y = 0 and ay = 0, the vertical component of velocity is also 0 m/s.
(d) The magnitude of the velocity (speed) when it hits the floor can be found using:
v = √(v0x^2 + v0y^2)
Since v0x = 2.5 m/s and v0y = 0, the magnitude of velocity is:
v = √(2.5^2 + 0) = √6.25 = 2.5 m/s
(e) The angle of impact below the horizontal can be found using the equation:
θ = tan^(-1)(v0y / v0x)
Since v0y = 0 and v0x = 2.5 m/s, the angle of impact is:
θ = tan^(-1)(0 / 2.5) = tan^(-1)(0) = 0 degrees
(f) The expression for the height of the object (y) in terms of x, v0x, y0, and g is:
y = y0 - (g * x^2) / (2 * v0x^2)
To solve parts (b) to (f), we need to consider the projectile motion of the object.
Let's start by evaluating the horizontal motion of the object (x-axis):
(b) The horizontal distance the object lands from the edge of the tabletop can be determined using the formula:
d = v0x * t,
where d is the horizontal distance and t is the time of flight.
To find t, we need to determine the time it takes for the object to hit the ground. We can use the vertical motion of the object to find this time.
Next, let's analyze the vertical motion of the object (y-axis):
(c) The vertical component of velocity when the object hits the ground can be found using the equation:
vfy = v0y + gt,
where vfy is the final vertical velocity, v0y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight.
To find v0y, we need to determine the initial vertical velocity. Since the table is at a height of y0 = 0.55 m above the ground, we can use the equation:
v0y = √(2g * y0).
(d) The magnitude of the velocity (speed) when it hits the floor can be found using the formula:
vf = √(vfx² + vfy²),
where vf is the final velocity, vfx is the final horizontal velocity, and vfy is the final vertical velocity.
(e) The angle of impact (θ) below the horizontal can be determined using the formula:
θ = atan(vfy / vfx),
where atan is the inverse tangent function.
(f) The height (y) of the object can be expressed in terms of x, v0x, y0, and g using the equation of motion for vertical motion:
y = y0 + x * tan(θ) - (gx²) / (2v0x²).
Now, let's calculate the individual values step by step, using the given information.
vertical position y(t) = 0.55 - 4.9t^2
horizontal position x(t) = 2.5t
use y to find how long it takes to fall to the floor (y=0)
vertical speed vy(t) = -9.8t
horizontal speed vx(t) = 2.5
(e) tanθ = vy/vx
The rest is just Algebra I. What do you get?