Question

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

Answer 1

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.