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 10.1 m/s at an angle of 62.4° 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?

Question

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 10.1 m/s at an angle of 62.4° 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?

I got the initial velocity of y to be 8.87m/s; however, I don't know the next steps to take.

Please help! Thanks

Answer 1

Vo = 10.1m/s[62.4o].

Yo = 10.1*sin62.4 = 8.95 m/s.

Y^2 = Yo^2 + 2g*h.
Y = 0, g = -9.8 m/s^2, h = ?.

Answer 2

To solve this problem, you can start by breaking down the initial velocity of the first aid kit into its horizontal (Vx) and vertical (Vy) components.

Given that the initial velocity of the kit is 10.1 m/s at an angle of 62.4° above the horizontal, you can calculate the vertical component (Vy) using trigonometry:

Vy = V * sin(θ)
= 10.1 m/s * sin(62.4°)
≈ 8.87 m/s

Since the vertical speed of the kit is zero when it is caught, you can use the equation of motion to find the time it takes for the kit to reach that point:

Vy = g * t
0 = 9.8 m/s^2 * t
t = 0 s

This means the kit takes 0 seconds to reach the point of being caught, which suggests that the two climbers are at the same height. Therefore, the vertical height difference between the two climbers is 0 meters.

This result implies that the two climbers are at the same height at the time of the catch, or the problem might have some additional information missing or incorrect. Double-check the problem statement for any possible errors.