Questions: Problem 9: (8% of Assignment Value) A basketball center fires a free-throw with initial velocity (v0=9.5 mathrm~m / mathrms) at an angle (theta=26^circ) above the horizontal. Use a Cartesian coordinate system with the origin located at the position the ball was released, with the ball's horizontal velocity in the positive x direction and vertical component in the positive y-direction. Assume the basketball encounters no air resistance. - Part (a) Determine the maximum vertical height (hmax), in meters, the ball attains above the release point. (hmax=0.8800 mathrm~m) Part (b) Determine the time, (t) in seconds, the basketball takes to reach its maximum vertical height. (t=)

Problem 9: (8% of Assignment Value) A basketball center fires a free-throw with initial velocity (v0=9.5 mathrm~m / mathrms) at an angle (theta=26^circ) above the horizontal. Use a Cartesian coordinate system with the origin located at the position the ball was released, with the ball's horizontal velocity in the positive x direction and vertical component in the positive y-direction. Assume the basketball encounters no air resistance. - Part (a) Determine the maximum vertical height (hmax), in meters, the ball attains above the release point. (hmax=0.8800 mathrm~m) Part (b) Determine the time, (t) in seconds, the basketball takes to reach its maximum vertical height. (t=)
Transcript text: Problem 9: ( $8 \%$ of Assignment Value) A basketball center fires a free-throw with initial velocity $v_{0}=9.5 \mathrm{~m} / \mathrm{s}$ at an angle $\theta=26^{\circ}$ above the horizontal. Use a Cartesian coordinate system with the origin located at the position the ball was released, with the ball's horizontal velocity in the positive x direction and vertical component in the positive y -direction. Assume the basketball encounters no air resistance. - Part (a) Determine the maximum vertical height $h_{\max }$, in meters, the ball attains above the release point. \[ h_{\max }=0.8800 \mathrm{~m} \] Part (b) Determine the time, $t$ in seconds, the basketball takes to reach its maximum vertical height. \[ t= \]
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Solution

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Solution Steps

Step 1: Identify the Vertical Component of Initial Velocity

The initial velocity \( v_0 = 9.5 \, \text{m/s} \) is given at an angle \( \theta = 26^\circ \). We need to find the vertical component of this velocity, \( v_{0y} \), using the formula: \[ v_{0y} = v_0 \sin \theta \] Substituting the given values: \[ v_{0y} = 9.5 \sin(26^\circ) \]

Step 2: Calculate the Vertical Component of Initial Velocity

Calculate \( v_{0y} \) using the sine function: \[ v_{0y} = 9.5 \times 0.4384 = 4.1648 \, \text{m/s} \]

Step 3: Use Kinematic Equation to Find Time to Maximum Height

The time to reach maximum height occurs when the vertical velocity is zero. Using the kinematic equation: \[ v_{y} = v_{0y} - gt \] Set \( v_{y} = 0 \) and solve for \( t \): \[ 0 = 4.1648 - 9.81t \] \[ t = \frac{4.1648}{9.81} \]

Step 4: Calculate the Time to Maximum Height

Perform the division to find \( t \): \[ t = \frac{4.1648}{9.81} \approx 0.4245 \, \text{s} \]

Final Answer

The time the basketball takes to reach its maximum vertical height is \(\boxed{0.4245 \, \text{s}}\).

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