Question

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 5.30 m/s from a height h = 1.20 m. Marble 2 is launched from ground level with a speed of 7.50 m/s at an angle = 45.0° above the horizontal. Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2 ? Give the x and y coordinates. [1​

Answer 1

The x-coordinate of the collision point is approximately 2.744 m and the y-coordinate is approximately 1.958 m.

Step-by-step explanation:

The marbles collide at the same height as marble 1. To find the x-coordinate, we can use the equation:

x = (v1x * t) = (5.3 m/s * t)

To find the y-coordinate, we can use the equation:

y = (h + v2y * t - (1/2) * g * t^2) = (1.2 m + (7.5 m/s * sin(45°)) * t - (1/2) * 9.8 m/s^2 * t^2)

Substituting the values:

x = (5.3 m/s * t)

y = (1.2 m + (7.5 m/s * 0.707) * t - (4.9 m/s^2 * t^2))

Set the x-coordinate equation equal to the y-coordinate equation:

5.3 * t = 1.2 + 7.5 * 0.707 * t - 4.9 * t^2

This equation can be rearranged to:

4.9 * t^2 - 2.715 * t + 1.2 = 0

Using the quadratic formula, we can solve for t:

t = (2.715 ± √(2.715^2 - 4 * 4.9 * 1.2)) / (2 * 4.9)

t = (2.715 ± √(7.366225 - 23.76)) / 9.8

t ≈ (-0.785) or t ≈ 0.518

Since time cannot be negative, t ≈ 0.518 seconds.

Now, substitute the value of t back into the x-coordinate equation:

x = (5.3 m/s * 0.518 s) ≈ 2.744 m

The x-coordinate of the collision point is approximately 2.744 m.

Substitute the value of t back into the y-coordinate equation:

y = (1.2 m + (7.5 m/s * sin(45°)) * 0.518 s - (1/2) * 9.8 m/s^2 * (0.518 s)^2)

y ≈ 1.958 m

The y-coordinate of the collision point is approximately 1.958 m.