Question

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

Answer:

15.1°

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

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

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:

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