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 12.8 m/s at an angle of 65.7° 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 difference between the two climbers?
recall that the max height is
(v sinθ)^2 / (2g)
To find the vertical height difference between the two climbers, we need to analyze the motion of the first aid kit. From the given information, we have the initial velocity (12.8 m/s) and the angle above the horizontal (65.7°).
First, let's break down the initial velocity into its vertical and horizontal components. The vertical component can be found using the equation:
Vertical component = initial velocity * sin(angle)
Vertical component = 12.8 m/s * sin(65.7°)
Next, we need to find the time taken for the first aid kit to reach its highest point. At the highest point, the vertical velocity will be zero. We can use the vertical component of the initial velocity and the acceleration due to gravity (-9.8 m/s^2) to find this time using the equation:
Vertical velocity = initial vertical velocity + acceleration * time
0 m/s = (12.8 m/s * sin(65.7°)) + (-9.8 m/s^2) * time
Now, solve for time.
0 = (12.8 m/s * sin(65.7°)) - 9.8 m/s^2 * time
9.8 m/s^2 * time = (12.8 m/s * sin(65.7°))
time = (12.8 m/s * sin(65.7°)) / 9.8 m/s^2
Once we have the time taken to reach the highest point, we can use the equation for vertical displacement to find the vertical height difference between the two climbers:
Vertical height difference = vertical component * time + (1/2) * acceleration * time^2
Vertical height difference = (12.8 m/s * sin(65.7°)) * time + (1/2) * (-9.8 m/s^2) * time^2
Substitute the value of time that we found earlier and calculate the vertical height difference to get the final answer.