Question

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the acceleration due to gravity is 9.81 m/s 2. The puck experiences no significant air resistance as it falls. The height of the platform 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

Answer 1

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

$\theta=tan^{-1}(\frac{v_y}{v_x})$       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

$v_x=v_o$

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

$v_y^2=v_{oy}^2+2gy$         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

$v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}$

Finally, you use the equation (1) to find the angle:

$\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°$

The angle below the horizontal is 12.60°