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 the final velocity and the angle the puck has the moment it hits the ground
To determine the final velocity and the angle at which the puck hits the ground, we can use the equations of motion.
The first step is to determine the time it takes for the puck to fall from the horizontal desk to the ground.
Using the equation:
h = (1/2)gt^2
where h is the height (1.2m) and g is the acceleration due to gravity (9.8 m/s^2), we can rearrange the equation to solve for t:
t = sqrt(2h / g)
= sqrt((2 * 1.2) / 9.8)
= 0.497 seconds (rounded to 3 decimal places)
Next, we can determine the horizontal component of the velocity, vx.
vx = distance / time = 0.0m / 0.497s = 0.0 m/s
Now, we can determine the vertical component of the velocity, vy, just before hitting the ground.
Using the equation:
vy = u sin(theta) - gt
where u is the initial speed (1.5 m/s), theta is the angle the puck is launched at, and g is the acceleration due to gravity (9.8 m/s^2), we can rearrange the equation to solve for vy:
vy = u sin(theta) - gt
vy + gt = u sin(theta)
(-9.8 m/s^2)(0.497 s) = (1.5 m/s) sin(theta) [substituting values]
-4.861 m/s = 1.5 sin(theta)
sin(theta) = -4.861 / 1.5
theta = arcsin(-4.861 / 1.5)
theta ≈ -66.17 degrees (rounded to 2 decimal places)
Note: The negative sign indicates that the puck is projected downwards.
Finally, we can determine the final velocity, v, just before hitting the ground.
Using the equation:
v = sqrt(vx^2 + vy^2)
v = sqrt((0.0 m/s)^2 + (-4.861 m/s)^2)
≈ 4.861 m/s (rounded to 3 decimal places)
Therefore, the final velocity of the puck just before hitting the ground is approximately 4.861 m/s, and the angle at which the puck hits the ground is approximately -66.17 degrees.
To determine the final velocity and angle of the puck when it hits the ground, we can use the principles of projectile motion.
First, let's calculate the time it takes for the puck to fall from the desk to the ground. We can use the equation for vertical displacement:
Δy = V₀y * t + (1/2) * g * t²
Where:
Δy = vertical displacement (1.2m)
V₀y = initial vertical velocity (in this case, since the puck is sliding horizontally, it is 0 m/s)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time
Rearranging the equation, we get:
1.2 = 0 * t + (1/2) * 9.8 * t²
1.2 = 4.9 * t²
t² = 1.2 / 4.9
t ≈ 0.274 s
Next, let's calculate the horizontal component of the puck's velocity. Since there are no horizontal forces acting on the puck, its horizontal velocity remains constant throughout the motion.
V₀x = initial horizontal velocity (1.5 m/s)
Now, let's determine the final velocity of the puck when it hits the ground. We can use the equations for horizontal and vertical velocity:
Vf = √(Vfx² + Vfy²)
Where:
Vf = final velocity
Vfx = final horizontal velocity
Vfy = final vertical velocity
Since there is no horizontal acceleration, Vfx = V₀x = 1.5 m/s.
For vertical velocity, we can use the equation:
Vfy = V₀y + g * t
Vfy = 0 + 9.8 * 0.274
Vfy ≈ 2.68 m/s
Now, let's calculate the final velocity:
Vf = √(Vfx² + Vfy²)
Vf = √(1.5² + 2.68²)
Vf ≈ 3.10 m/s
Finally, let's calculate the angle at which the puck hits the ground. We can use the equation for the angle of the projectile motion:
θ = tan⁻¹(Vfy / Vfx)
θ = tan⁻¹(2.68 / 1.5)
θ ≈ 59.3°
Therefore, the final velocity of the puck when it hits the ground is approximately 3.10 m/s, and the angle it makes with the horizontal at that moment is approximately 59.3°.
To determine the final velocity and the angle at which the puck hits the ground, we need to analyze the projectile motion of the puck.
First, let's break down the problem into horizontal and vertical components:
1. Horizontal motion: The horizontal velocity remains constant throughout the motion as there are no horizontal forces acting on the puck. Therefore, the horizontal component of the velocity (Vx) remains constant, and the horizontal displacement (dx) can be calculated using the formula: dx = Vx * t, where t is the time of flight.
2. Vertical motion: The vertical motion of the puck is affected by the acceleration due to gravity. The vertical component of the velocity (Vy) changes as the puck falls, and the vertical displacement (dy) can be calculated using the formula: dy = Vy * t + (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's proceed to calculate the final velocity and the angle at which the puck hits the ground:
1. Determine the time of flight (t):
To find the time of flight, we can use the formula for vertical displacement and solve for t. In this case, dy is equal to the height from which the puck is dropped, i.e. 1.2m. Therefore, the equation becomes:
1.2m = Vy * t + (1/2) * g * t^2.
2. Solve for Vy:
At the highest point of the trajectory, the vertical velocity becomes 0 since the puck momentarily stops before falling. Therefore, we can use the following equation to find Vy:
0 = Vy + g * t.
From equation 2, we can solve for t and substitute it back into equation 1 to solve for Vy.
3. Calculate Vx:
The horizontal velocity (Vx) remains constant throughout the motion. Therefore, the initial horizontal velocity (Vx_initial) is equal to the final horizontal velocity (Vx_final). Given that the initial speed of the puck is 1.5 m/s, Vx_initial = Vx_final = 1.5 m/s.
4. Determine the final velocity:
The final velocity (V) is the resultant of the horizontal and vertical velocities. We can calculate it using the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2).
5. Calculate the angle (θ):
The angle at which the puck hits the ground can be determined using the tangent function:
θ = arctan(Vy / Vx).
By following these steps, you should be able to determine the final velocity and the angle at which the puck hits the ground in this experiment.