A child uses a rubber band to launch a bottle cap at an angle of 35.5° above the horizontal. The cap travels a horizontal distance of 1.30 m in 1.60 s. What was the initial speed of the bottle cap, just after leaving the rubber band?
3.67
To find the initial speed of the bottle cap, we can use the projectile motion equations. The horizontal and vertical components of motion are independent of each other.
First, let's find the initial vertical velocity of the bottle cap. We know that the angle of launch is 35.5° above the horizontal. We can use this angle to determine the vertical component of velocity.
1. Start by creating a right triangle with the given angle of 35.5°.
2. The vertical velocity component can be determined using the equation: Velocity (V) = Initial Velocity (Vi) * sin(θ), where θ is the launch angle.
3. Rearrange the equation to solve for Vi: Vi = V / sin(θ).
Next, let's find the time of flight for the bottle cap. We know that it travels a horizontal distance of 1.30 m and takes 1.60 s.
1. Use the equation: Distance (d) = Velocity (V) * Time (t) to find the horizontal velocity component. In this case, the distance is the horizontal distance traveled and the time is the time of flight.
2. Rearrange the equation to solve for V: V = d / t.
Now that we have both the vertical and horizontal velocity components, we can find the initial speed (V0) using the Pythagorean theorem:
1. Use the equation: V0 = sqrt(V_horizontal^2 + V_vertical^2), where V_horizontal is the horizontal velocity component and V_vertical is the vertical velocity component.
Let's plug in the values:
Vi = V / sin(θ) = V / sin(35.5°)
V = d / t = 1.30 m / 1.60 s
Now, calculate the initial vertical velocity:
Vi_vertical = V / sin(35.5°)
Finally, find the initial speed of the bottle cap:
V0 = sqrt(V_horizontal^2 + V_vertical^2)
Substituting the values, you can calculate the initial speed of the bottle cap just after leaving the rubber band.