Question

A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height of the tabletop above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground? A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height of the tabletop above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground? 72.6° 31.8° 15.1° 77.2° 22.8°

Answer 1

Using the kinematic equations for projectile motion, the final angle below the horizontal of the velocity of the puck just before it hits the ground is calculated to be 31.8° based on the given height of the fall and the initial horizontal velocity.

Step-by-step explanation:

To find the angle below the horizontal of the velocity of the puck just before it hits the ground, we can use the kinematic equations for projectile motion. Considering that the puck experiences no air resistance, its horizontal velocity component remains constant at 23.2 m/s, and the vertical velocity component increases due to gravity (9.81 m/s2). First, we calculate the time it takes for the puck to fall 2.00 m using the vertical motion equation:

h = v0yt + (1/2)gt2

Since the puck slides off the table, the initial vertical velocity v0y is 0, so the equation simplifies to:

h = (1/2)gt2

Solving for t, we get:

t = sqrt((2h)/g)

Now that we have the time of fall, we can find the vertical velocity component just before impact using:

vy = gt

Finally, the angle θ can be calculated using the vertical and horizontal components:

tan(θ) = vy/vx

Plugging in the values:

θ = arctan(vy/vx)

After performing the calculations with the given numbers, we find that the angle is 31.8° below the horizontal just before the puck hits the ground. Therefore, the correct answer is 31.8°.

Answer 2

15.1°

Step-by-step explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

`v_x = 23.2 m/s`

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

`v_y(t)= v_(y0) -gt` (1)

where

`v_(y0)=0` is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

`h=(1)/(2)gt^2\\t=\sqrt{(2h)/(g)}=\sqrt{(2(2.00 m))/(9.8 m/s^2)}=0.64 s`

Substituting t into (1) we find the final vertical velocity

`v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s`

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

`tan \theta = (|v_y|)/(v_x)=(6.3 m/s)/(23.2 m/s)=0.272\\\theta = tan^(-1) (0.272)=15.1^(\circ)`