A spring cannon is located at the edge of a table that is 1.70 m above the floor. A steel ball is launched from the cannon with speed v0 at 25.0° above the horizontal.
(a) Find the horizontal displacement component of the ball to the point where it lands on the floor as a function of v0. We write this function as x(v0).
To find the horizontal displacement component of the ball, we can use the equations of motion for projectile motion.
First, let's analyze the motion in the x-direction. Since there are no horizontal forces acting on the ball, its horizontal velocity remains constant throughout the motion. Therefore, the horizontal displacement can be expressed as:
x = v0 * cos(θ) * t
where:
- v0 is the initial velocity of the ball
- θ is the angle of projection (25.0° in this case)
- t is the time taken for the ball to hit the floor
To find the time t, we need to analyze the motion in the y-direction. We can use the equation:
y = y0 + v0 * sin(θ) * t - (1/2) * g * t^2
where:
- y0 is the initial vertical position (1.70 m in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
The ball hits the floor when y becomes zero. Substituting y = 0 into the equation above, we can solve for t:
0 = y0 + v0 * sin(θ) * t - (1/2) * g * t^2
=> 0 = 1.70 + (v0 * sin(25°)) * t - (4.9) * t^2
This equation is a quadratic equation in terms of t. We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / 2a
where:
- a = -4.9
- b = v0 * sin(25°)
- c = 1.70
Once we have found the value of t, we can substitute it back into the expression for x to find the horizontal displacement component x(v0).
Therefore, the function x(v0) is given by:
x(v0) = v0 * cos(θ) * t
where t is calculated using the quadratic formula mentioned above.