A physics book slides off a horizontal tabletop with a speed of 1.10m/s{\rm m}/{\rm s} . It strikes the floor in 0.350s{\rm s} . Ignore air resistance.
Assuming question wants height of table, Δy.
Use kinematics equation:
Δy = viyΔt + (1/2)gΔt²
Assuming 1.1m/s is horizontal velocity (slides off table), so viy=0
g=-9.8 m/s²
Δy=0+(1/2)(-9.8)(0.35²)
=-4.9(0.35²)
=-0.060 m
Floor is 0.06 m below table (not quite a table, usually 0.75 m high)
Horizontal distance covered:
vix*Δt
=1.1*0.35
=0.385 m.
To solve this problem, we can use the equations of kinematics to find the distance traveled by the book before hitting the floor.
First, let's identify the given information:
Initial velocity (u) = 1.10 m/s
Time (t) = 0.350 s
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming we are on Earth and ignoring air resistance)
Now, we can use the equation of motion:
s = ut + (1/2)gt^2
where:
s = distance traveled
u = initial velocity
t = time
g = acceleration due to gravity
Plugging in the values, we get:
s = (1.10 m/s)(0.350 s) + (1/2)(9.8 m/s^2)(0.350 s)^2
Simplifying:
s = 0.385 m + (1/2)(9.8 m/s^2)(0.1225 s^2)
s = 0.385 m + 0.6 m
s = 0.985 m
Therefore, the physics book slides a distance of 0.985 meters before hitting the floor.
To solve this problem, we can use the following kinematic equations:
1. The equation for displacement in terms of initial velocity, time, and acceleration:
Δx = v₀t + (1/2)at²
2. The equation for final velocity in terms of initial velocity, acceleration, and time:
v = v₀ + at
3. The equation for displacement in terms of initial and final velocities, and acceleration:
Δx = (v² - v₀²)/(2a)
Here, we are given:
Initial velocity, v₀ = 1.10 m/s
Time, t = 0.350 s
First, we can find the acceleration using equation 2:
v = v₀ + at
We know v = 0 m/s since the book comes to a stop when it reaches the ground. Plugging in the values, we get:
0 = 1.10 + a * 0.350
Simplifying the equation:
0.350a = -1.10
a = -1.10/0.350
a = -3.14 m/s² (rounded to two decimal places)
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
Next, we can find the displacement using equation 1:
Δx = v₀t + (1/2)at²
Plugging in the values, we get:
Δx = 1.10 * 0.350 + (1/2) * (-3.14) * (0.350)²
Δx = 0.385 + (-0.06475)
Δx = 0.32025 m (rounded to five decimal places)
Therefore, the displacement of the book when it strikes the ground is 0.32025 meters.