A basketball is shot from an initial height of 2.4 m with an initial speed (v0 = 12 m/s) directed at an angle (35 degrees) above the horizontal.
Additional information:
Height of the basket is 3.05 m
To determine whether the basketball will make it into the basket, we need to calculate its horizontal distance traveled when it reaches a vertical height equal to the height of the basket.
The first step is to break down the initial velocity into its horizontal and vertical components. Since the basketball is shot at an angle above the horizontal, we can use trigonometry to find these components.
The horizontal component, Vx, can be found using the equation: Vx = v0 * cos(theta), where v0 is the initial speed and theta is the angle. Substituting the given values, we get Vx = 12 * cos(35 degrees).
The vertical component, Vy, can be found using the equation: Vy = v0 * sin(theta), where v0 is the initial speed and theta is the angle. Substituting the given values, we get Vy = 12 * sin(35 degrees).
We can use the following kinematic equation to find the time it takes for the basketball to reach the same elevation as the basket: y = y0 + Vy * t - (1/2) * g * t^2, where y is the final height (3.05 m), y0 is the initial height (2.4 m), Vy is the vertical component of the initial velocity (12 * sin(35 degrees)), t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
Rearranging the equation, we get: (1/2) * g * t^2 - Vy * t + (y0 - y) = 0. This is a quadratic equation with respect to time. We can solve for t using the quadratic formula: t = (-b +/- sqrt(b^2 - 4ac))/(2a), where a = (1/2) * g, b = -Vy, and c = (y0 - y).
Once we have the time it takes for the basketball to reach the same elevation as the basket, we can calculate the horizontal distance traveled using the equation: x = Vx * t.
Now, let's substitute the given values into the formulas to find the answer.