physics students are performing an experiment and slide a hockey puck off a horizontal desk that is 1.2m high. the initial speed of the puck is 1.5m/s. determine how far from the base of the desk the puck will land.
To solve this problem, we need to use the following kinematic equation:
\[ d = v_{\text{initial}} t + \frac{1}{2} a t^2 \]
We need to find the time it takes for the puck to hit the ground, so we can set the equation equal to the vertical distance the puck falls, which is the height of the desk:
\[ 1.2 \, \text{m} = -(\text{acceleration due to gravity}) \cdot t^2 \]
This is because the only force acting on the puck in the vertical direction is the acceleration due to gravity directed downward. The negative sign indicates that the displacement is in the opposite direction to the acceleration. So, the acceleration due to gravity (g) will be negative.
The initial velocity of the puck in the vertical direction is zero since it is sliding horizontally off the desk, and there are no vertical upward forces.
Next, let's find the time it takes for the puck to hit the ground:
\[ 1.2 \, \text{m} = -\frac{1}{2} g \cdot t^2 \]
Rearranging the equation:
\[ t^2 = \frac{2 \cdot 1.2 \, \text{m}}{g} \]
Simplifying:
\[ t = \sqrt{\frac{2 \cdot 1.2 \, \text{m}}{g}} \]
Now, let's find the horizontal distance the puck travels before hitting the ground. Since there is no horizontal acceleration, the horizontal speed remains constant. Hence,
\[ \text{distance} = \text{speed} \times \text{time} \]
Substituting the given values,
\[ \text{distance} = (1.5 \, \text{m/s}) \times \sqrt{\frac{2 \cdot 1.2 \, \text{m}}{g}} \]
Now, let's compute the value using the standard acceleration due to gravity:
\[ \text{distance} \approx (1.5 \, \text{m/s}) \times \sqrt{\frac{2 \cdot 1.2 \, \text{m}}{9.8 \, \text{m/s}^2}} \]
Simplifying the equation:
\[ \text{distance} \approx (1.5 \, \text{m/s}) \times \sqrt{\frac{2 \cdot 1.2 \, \text{m}}{9.8 \, \text{m/s}^2}} \approx 0.81 \, \text{m} \]
Therefore, the puck will land approximately 0.81 meters from the base of the desk.
To determine how far from the base of the desk the puck will land, you can use the equations of motion to find the horizontal distance traveled by the puck.
The vertical distance traveled by the puck when it falls from the desk is given by the equation:
h = 0.5 * g * t^2
Where h is the vertical distance (1.2 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken to fall.
Rearranging the equation to solve for t:
t^2 = (2h) / g
t = sqrt((2h) / g)
Substituting the given values:
t = sqrt((2 * 1.2) / 9.8)
t ≈ sqrt(0.2449)
t ≈ 0.4949 s
Now, to find the horizontal distance traveled by the puck, we can use the equation:
d = v * t
Where d is the horizontal distance, v is the initial horizontal velocity of the puck (1.5 m/s), and t is the time taken to fall (0.4949 s).
Substituting the given values:
d = 1.5 * 0.4949
d ≈ 0.7424 m
Therefore, the puck will land approximately 0.7424 meters from the base of the desk.
To determine the horizontal distance from the base of the desk where the puck will land, we can use the equations of motion and the principles of projectile motion. Let's break down the problem into its components:
1. Initial vertical velocity (v₀y): The puck starts from rest vertically, so its initial vertical velocity is zero (v₀y = 0).
2. Final vertical velocity (vᵢy): We need to determine the final vertical velocity when the puck hits the ground. We can use the equation:
vᵢy² = v₀y² + 2aΔy,
where v₀y is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s²), and Δy is the vertical distance traveled. Since the puck falls downward, we have Δy = -1.2 m. Therefore:
vᵢy² = 0² + 2(-9.8)(-1.2).
Solving for vᵢy gives:
vᵢy = √(2(9.8)(1.2)) = 4.077 m/s (rounded to three decimal places).
3. Total time of flight (t): We can use the equation:
Δy = v₀y * t + (1/2) * a * t²,
where Δy is the vertical distance traveled and a is the acceleration due to gravity (-9.8 m/s²). Solving for t:
-1.2 = 0 * t + (1/2) * (-9.8) * t²,
-1.2 = -4.9t².
Dividing both sides by -4.9 gives:
0.2449 = t².
Taking the square root of both sides, we get:
t = √0.2449 = 0.495 s (rounded to three decimal places).
4. Horizontal distance (d): We can use the equation:
d = v₀x * t,
where d is the horizontal distance, v₀x is the initial horizontal velocity, and t is the total time of flight. Since there are no horizontal forces acting on the puck, its velocity remains constant throughout. Consequently, the initial horizontal velocity is the same as the final horizontal velocity. Therefore:
v₀x = vᵢx.
Assuming there is no air resistance, the initial horizontal velocity (v₀x) can be calculated using:
v₀x = vᵢx = 1.5 m/s (given).
Substituting the values into the equation, we get:
d = (1.5 m/s) * (0.495 s) = 0.7425 m (rounded to four decimal places).
Therefore, the puck will land approximately 0.7425 meters from the base of the desk.