A rock climber throws a small first aid kit to another climber who is higher up the mountain. The initial velocity of the kit is 17 m/s at an angle of 50° above the horizontal. At the instant when the kit is caught, it is traveling horizontally, so its vertical speed is zero. What is the vertical height between the two climbers?

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To find the vertical height between the two climbers, we can use the equations of motion and divide the problem into horizontal and vertical components.

First, let's break down the initial velocity into its horizontal and vertical components. The initial velocity is 17 m/s at an angle of 50° above the horizontal. We can find the horizontal component by multiplying the initial velocity by the cosine of the angle, and the vertical component by multiplying the initial velocity by the sine of the angle.

Horizontal component: 17 m/s * cos(50°) = 17 m/s * 0.6428 ≈ 10.93 m/s
Vertical component: 17 m/s * sin(50°) = 17 m/s * 0.766 ≈ 13.03 m/s

Since the vertical speed becomes zero when the kit is caught, we can use the vertical component to find the time of flight, which is the time it takes for the kit to reach its highest point before coming back down.

Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can plug in the values:

0 m/s = 13.03 m/s + (-9.8 m/s^2) * t

Solving this equation for t, we get:

t ≈ 1.33 seconds

Now, we can use the time of flight to find the maximum height reached by the kit.

Using the equation s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time, we can plug in the values:

s = 13.03 m/s * 1.33 seconds + (1/2) * (-9.8 m/s^2) * (1.33 seconds)^2

Simplifying this equation, we get:

s ≈ 8.66 meters

Therefore, the vertical height between the two climbers is approximately 8.66 meters.