A glass is slid across a table top, goes of the edge and falls to the floor below. The table is 0.860 m high and the glass lands 1.77 m from the edge.
(a) With what velocity (m/s) did the glass leave the table top?
(b) What was the direction of the glass's velocity (degrees clockwise from the horizontal)as it hit the ground?
To solve this problem, we can use the equations of motion and the principles of projectile motion. Let's start by breaking down the problem and finding the necessary information.
Given:
- Height of the table (h) = 0.860 m
- Horizontal distance from the edge (range) = 1.77 m
(a) With what velocity (m/s) did the glass leave the table top?
To solve for the velocity, we can consider the vertical motion of the glass. We'll use the equation of motion:
h = (1/2)gt^2
where:
- h is the height (0.860 m)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time of flight
First, let's calculate the time of flight. At the highest point, the vertical velocity is zero. We can use the equation:
v = u + gt
where:
- v is the final velocity (0 m/s at the highest point)
- u is the initial velocity (what we want to find)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time of flight (what we want to find)
Rearranging the equation, we have:
t = (v - u) / g
Since the final velocity is zero at the highest point, the equation becomes:
t = -u / g
Now, let's substitute this value of t into the equation of motion for vertical motion:
h = (1/2)gt^2
0.860 = (1/2)g (-u/g)^2
0.860 = (1/2)u^2/g
Simplifying the equation, we have:
u^2 = 2gh
Plugging in the values:
u = √(2 × 9.8 × 0.860)
u ≈ 4.17 m/s
Therefore, the glass left the table top with a velocity of approximately 4.17 m/s.
(b) What was the direction of the glass's velocity (degrees clockwise from the horizontal) as it hit the ground?
To find the angle with the horizontal, we can use trigonometry and the range of the projectile. The range equation is given by:
R = (u^2 × sin(2θ)) / g
where:
- R is the range (1.77 m)
- u is the initial velocity (4.17 m/s)
- θ is the angle with the horizontal (what we want to find)
- g is the acceleration due to gravity (9.8 m/s^2)
Rearranging this equation, we can solve for θ:
θ = (1/2) × arcsin((R × g) / u^2)
Plugging in the values:
θ = (1/2) × arcsin((1.77 × 9.8) / (4.17)^2)
θ ≈ 24.7 degrees
Therefore, the glass's velocity made an angle of approximately 24.7 degrees clockwise from the horizontal as it hit the ground.
Note: The negative angle indicates the clockwise direction from the horizontal.