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. 


To find the time, \( \Delta t \), it takes for a ball to fall from a table of height \( h \) with an initial velocity \( v_i \) at an angle \( \theta \) to the horizontal, and under the influence of vertical acceleration \( a_y = -g \), 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:
   \[
   v_{i_y} = v_i \sin \theta
   \]
   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:
   \[
   y(t) = y_0 + v_{i_y} t + \frac{1}{2} a_y t^2
   \]
   Here, \( y_0 = h \) is the initial height of the ball, and the final vertical position is \( y(t) = 0 \) (since the ball reaches the floor).

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

3. **Solving for time \( \Delta t \)**:  
   We now solve this quadratic equation for \( t \):
   \[
   0 = h + v_i \sin \theta \cdot t - \frac{1}{2} g t^2
   \]
   Rearranging:
   \[
   \frac{1}{2} g t^2 - v_i \sin \theta \cdot t - h = 0
   \]
   This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:
   - \( a = \frac{1}{2} g \)
   - \( b = -v_i \sin \theta \)
   - \( c = -h \)

   Using the quadratic formula:
   \[
   t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Substituting the values of \( a \), \( b \), and \( c \):
   \[
   t = \frac{-(-v_i \sin \theta) \pm \sqrt{(-v_i \sin \theta)^2 - 4 \cdot \frac{1}{2} g \cdot (-h)}}{2 \cdot \frac{1}{2} g}
   \]
   Simplifying the expression:
   \[
   t = \frac{v_i \sin \theta \pm \sqrt{(v_i \sin \theta)^2 + 2gh}}{g}
   \]

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

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

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

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Here are 5 related questions to explore further:
1. How would the time change if the angle \( \theta \) is increased?
2. What is the effect of height \( h \) 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 \( v_i \) is zero?
5. How would air resistance affect this motion?

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

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.

Explore how to calculate the time it takes for a ball to fall from a table of height h to the floor, given an initial velocity vi at an angle θ and acceleration due to gravity g. This problem covers concepts like projectile motion, vertical velocity components, and solving quadratic equations. Ideal for advanced high school physics students.

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