Math Problem Statement

Find the time, Δt, it takes the ball to fall from the table of height h to the floor with an initial velocity of vi at an angle of θ with respect to the horizontal, and a vertical acceleration of ay = -g. Your answer will be an equation (not a number) in terms of vi, θ, h and g.

Solution

To find the time, Δt\Delta t, it takes for a ball to fall from a table of height hh with an initial velocity viv_i at an angle θ\theta to the horizontal, and under the influence of vertical acceleration ay=ga_y = -g, we will focus on the vertical motion of the ball.

Step-by-step solution:

  1. Vertical component of the initial velocity:
    The initial velocity can be broken into two components: viy=visinθv_{i_y} = v_i \sin \theta This is the vertical component of the initial velocity.

  2. Equation of vertical motion:
    The general equation for the vertical position of the ball as a function of time is: y(t)=y0+viyt+12ayt2y(t) = y_0 + v_{i_y} t + \frac{1}{2} a_y t^2 Here, y0=hy_0 = h is the initial height of the ball, and the final vertical position is y(t)=0y(t) = 0 (since the ball reaches the floor).

    Substituting these values into the equation gives: 0=h+visinθt12gt20 = h + v_i \sin \theta \cdot t - \frac{1}{2} g t^2

  3. Solving for time Δt\Delta t:
    We now solve this quadratic equation for tt: 0=h+visinθt12gt20 = h + v_i \sin \theta \cdot t - \frac{1}{2} g t^2 Rearranging: 12gt2visinθth=0\frac{1}{2} g t^2 - v_i \sin \theta \cdot t - h = 0 This is a quadratic equation of the form at2+bt+c=0at^2 + bt + c = 0, where:

    • a=12ga = \frac{1}{2} g
    • b=visinθb = -v_i \sin \theta
    • c=hc = -h

    Using the quadratic formula: t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Substituting the values of aa, bb, and cc: t=(visinθ)±(visinθ)2412g(h)212gt = \frac{-(-v_i \sin \theta) \pm \sqrt{(-v_i \sin \theta)^2 - 4 \cdot \frac{1}{2} g \cdot (-h)}}{2 \cdot \frac{1}{2} g} Simplifying the expression: t=visinθ±(visinθ)2+2ghgt = \frac{v_i \sin \theta \pm \sqrt{(v_i \sin \theta)^2 + 2gh}}{g}

  4. Choosing the positive root:
    Since time tt must be positive, we take the positive root: Δt=visinθ+(visinθ)2+2ghg\Delta t = \frac{v_i \sin \theta + \sqrt{(v_i \sin \theta)^2 + 2gh}}{g}

Thus, the time Δt\Delta t it takes for the ball to fall to the floor is: Δt=visinθ+(visinθ)2+2ghg\boxed{\Delta t = \frac{v_i \sin \theta + \sqrt{(v_i \sin \theta)^2 + 2gh}}{g}}

Would you like more details on any step or have questions?


Here are 5 related questions to explore further:

  1. How would the time change if the angle θ\theta is increased?
  2. What is the effect of height hh on the time of flight?
  3. How would you find the horizontal distance the ball travels during this time?
  4. What happens to the time if the initial velocity viv_i is zero?
  5. How would air resistance affect this motion?

Tip: Always break down a 2D motion problem into vertical and horizontal components.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Quadratic Equations

Formulas

y(t) = y0 + v_{iy} t + 1/2 a_y t^2
Quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a
Vertical velocity component: v_{iy} = vi * sin(θ)

Theorems

Quadratic Formula
Equations of Motion under Constant Acceleration

Suitable Grade Level

Grade 11-12 (Advanced High School Physics)