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 16 m/s at an angle of 55° above the horizontal. The kit is caught right at the peak of the its trajectory. What is the vertical height between the two climbers?
Vo = 16m/s[55o]
Yo = 16*sin55 = 13.11 m/s.
Y^2 = Yo^2 + 2g*h
h = (Y^2-Yo^2)/2g
h = (0-13.11^2)/-19.6 = 8.77 m.
To find the vertical height between the two climbers, we can use the principles of projectile motion.
Step 1: Split the initial velocity into horizontal and vertical components.
The initial velocity of 16 m/s at an angle of 55° above the horizontal can be split into horizontal and vertical components.
Vertical component (Vy) = Initial velocity (V) × sine(angle)
Vertical component (Vy) = 16 m/s × sin(55°)
Vertical component (Vy) = 16 m/s × 0.819
Vertical component (Vy) ≈ 13.104 m/s
Step 2: Determine the time it takes for the kit to reach its peak.
The vertical motion is affected by gravity, which causes the object to decelerate until it reaches its peak and then accelerate downward. The time it takes to reach the peak is the same as the time it takes to fall from the peak to the catch point.
Using the equation: Δy = Vy × t + (1/2) × g × t², where Δy is the displacement in the vertical direction, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time.
At the peak, the vertical displacement (Δy) is equal to zero.
0 = 13.104 m/s × t + (1/2) × 9.8 m/s² × t²
Simplifying the equation:
0 = 13.104 m/s × t + 4.9 m/s² × t²
Rearranging the equation:
4.9 m/s² × t² + 13.104 m/s × t = 0
Using the quadratic formula: t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 4.9 m/s², b = 13.104 m/s, and c = 0.
t = (-13.104 m/s ± √((13.104 m/s)² - 4 × 4.9 m/s² × 0)) / (2 × 4.9 m/s²)
t ≈ -1.333 s (Ignoring negative value)
Therefore, the time it takes for the kit to reach its peak is approximately 1.333 seconds.
Step 3: Calculate the vertical displacement.
Using the equation: Δy = Vyt - (1/2) × g × t²
Substituting the known values:
Δy = 13.104 m/s × 1.333 s - (1/2) × 9.8 m/s² × (1.333 s)²
Δy ≈ 17.472 m - (1/2) × 9.8 m/s² × 1.777 m²
Δy ≈ 17.472 m - 8.866 m
Δy ≈ 8.606 m
Therefore, the vertical height between the two climbers is approximately 8.606 meters.
To find the vertical height between the two climbers, we need to determine the distance the first aid kit travels vertically before it reaches its peak height.
First, let's break down the initial velocity of the kit into its horizontal and vertical components. The horizontal component can be found by multiplying the initial velocity by the cosine of the angle:
Horizontal component = 16 m/s * cos(55°)
Similarly, the vertical component can be found by multiplying the initial velocity by the sine of the angle:
Vertical component = 16 m/s * sin(55°)
Now, the time taken for the kit to reach its peak can be calculated using the vertical component. We can use the equation:
Vertical component = (initial vertical velocity) * time + (acceleration due to gravity * time^2) / 2
Since the kit is caught right at the peak, its final vertical velocity is 0. Therefore, we can rearrange the equation to solve for time:
0 = (16 m/s * sin(55°)) * time - (9.8 m/s^2 * time^2) / 2
Now we can solve this quadratic equation to find the time it takes for the kit to reach its peak.
Once we have the time, we can use the formula for vertical distance traveled during constant acceleration to find the vertical height between the two climbers:
Vertical height = (initial vertical velocity * time) - (1/2 * acceleration due to gravity * time^2)
Using these calculations, we can find the vertical height between the two climbers.