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 21 m/s at an angle of 58° 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?
Easy. YOu know it went h height, with an initial vertical velocity of 21sin58.
Vf^2=Vi^2 + 2gh solve for h.
To find the vertical height between the two climbers, we need to determine the time it takes for the kit to reach its highest point and then fall back down to the horizontal level.
Step 1: Splitting the Velocity
The initial velocity of the kit has two components: the horizontal component (Vx) and the vertical component (Vy). We can find these components using trigonometry.
Vx = V * cos(θ)
Vy = V * sin(θ)
Given:
V = 21 m/s (initial velocity of the kit)
θ = 58° (angle above the horizontal)
Vx = 21 * cos(58°)
Vx = -10.65 m/s (Take the negative sign since the velocity is in the opposite direction of the positive x-axis)
Vy = 21 * sin(58°)
Vy = 17.77 m/s
Step 2: Finding the time to reach the highest point
To find the time it takes for the kit to reach its highest point, we need to consider the vertical motion, which is affected by gravity (g).
We can use the equation:
Vy = Voy + gt
Where:
Voy = initial vertical velocity (17.77 m/s)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time to reach the highest point
At the highest point, the vertical velocity becomes zero. Therefore, we can rewrite the equation as:
0 = 17.77 + (-9.8) * t
Solving for t:
-17.77 = -9.8 * t
t = -17.77 / -9.8
t = 1.81 seconds
Step 3: Calculate the Vertical Displacement
Now that we know the time it takes for the kit to reach the highest point, we can use the equation to find the vertical displacement (Δy) between the two climbers at that moment.
Δy = Voy * t + (1/2) * g * t²
Where:
Voy = initial vertical velocity (17.77 m/s)
t = time to reach the highest point (1.81 s)
g = acceleration due to gravity (approximately 9.8 m/s²)
Δy = 17.77 * 1.81 + (1/2) * 9.8 * 1.81²
Δy = 32.1 meters
Therefore, the vertical height between the two climbers is 32.1 meters.