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.

1 answer

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.