We are tasked with finding the initial velocity and the time it takes for the ball to reach the green when hit from the top of a mountain with a launch angle of 28.3° from the horizontal. Here's how we can solve it step-by-step:

Given dataWe are tasked with finding the initial velocity and the time it takes for the ball to reach the green when hit from the top of a mountain with a launch angle of 28.3° from the horizontal. Here's how we can solve it step-by-step:

Given

• Height of the mountain, $h = 400 \, \text{m}$
• Horizontal distance to the green, $d = 359 \, \text{m}$
• Launch angle, $\theta = 28.3^\circ$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

We need to find two things:

1. The initial velocity $v_0$
2. The time of flight $t$

Step 1: Split the motion into horizontal and vertical components

The horizontal and vertical motions can be treated separately using the following kinematic equations:

• Horizontal motion (constant velocity): $x = v_{0x} \cdot t$ where $v_{0x} = v_0 \cos \theta$, and $x = d$.

• Vertical motion (accelerated due to gravity): $y = v_{0y} \cdot t - \frac{1}{2} g t^2$ where $v_{0y} = v_0 \sin \theta$, and $y = -400 \, \text{m}$ (since the ball is falling 400 m down to the green).

Step 2: Horizontal motion equation

From the horizontal motion equation: $x = v_{0x} \cdot t$ $359 = (v_0 \cos 28.3^\circ) \cdot t$ This gives us the first equation: $t = \frac{359}{v_0 \cos 28.3^\circ}$ We will substitute this expression for $t$ into the vertical motion equation.

Step 3: Vertical motion equation

For the vertical motion, we use the equation: $y = v_{0y} \cdot t - \frac{1}{2} g t^2$ Substitute $y = -400 \, \text{m}$ and $v_{0y} = v_0 \sin 28.3^\circ$: $-400 = (v_0 \sin 28.3^\circ) \cdot t - \frac{1}{2} (9.8) t^2$ Now substitute the expression for $t$ from the horizontal equation: $t = \frac{359}{v_0 \cos 28.3^\circ}$ into the vertical equation to solve for $v_0$.

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Tip: In projectile problems, breaking down the motion into horizontal and vertical components is key to understanding the trajectory.