A monkey is hanging from a branch located exactly 24 meters along the ground from the hunter. The monkey is 14 meters above the ground. The hunter aims directly at the monkey and fires a water balloon at the monkey just as the monkey releases his grip. The balloon has an initial velocity of 40m/s.

A. How long until the balloon reaches the monkey
B. At what height above the ground will the monkey be hit by the balloon.
C. At what speed should the balloon be launched so that the monkey is hit by the balloon exactly 3 meters above the ground.

My answers: For A I got .695 seconds
For B I got 11.6 meters
For C I got 35.4 m/s 12.6 degrees above the ground.

Could you check my answers? Thank you!

To verify your answers, let's solve the problem step by step.

A. How long until the balloon reaches the monkey?

To solve this, we can analyze the horizontal and vertical motions of the water balloon. Since there is no horizontal acceleration (neglecting air resistance), the time it takes for the balloon to reach the monkey can be determined by dividing the horizontal distance by the horizontal velocity.

Given:
- Horizontal distance (x) = 24 meters
- Horizontal velocity (v_x) = 40 m/s

Using the formula: time (t) = distance (x) / velocity (v)
t = 24 m / 40 m/s
t = 0.6 seconds (not 0.695 seconds as you mentioned)

So, the correct answer for part A is 0.6 seconds.

B. At what height above the ground will the monkey be hit by the balloon?

To find the height, or vertical position, at which the monkey will be hit, we need to consider the vertical motion of the monkey and the water balloon.

Given:
- Initial vertical position (y_0) = 14 meters
- Vertical velocity (v_y) = 0 m/s (at the peak of the monkey's motion)
- Acceleration due to gravity (g) = 9.8 m/s²
- Time (t) = 0.6 seconds (from part A)

Using the formula: displacement (y) = initial position (y_0) + (velocity * time) + (0.5 * acceleration * time²)

y = 14 m + (0 m/s * 0.6 s) + (0.5 * 9.8 m/s² * (0.6 s)²
y ≈ 14 m + 0 m + 1.764 m
y ≈ 15.764 meters (not 11.6 meters as you mentioned)

Therefore, the correct answer for part B is approximately 15.764 meters.

C. At what speed should the balloon be launched so that the monkey is hit by the balloon exactly 3 meters above the ground?

In this case, we need to calculate the initial velocity (v_i) at which the balloon should be launched.

Given:
- Initial vertical position (y_0) = 14 meters
- Final vertical position (y) = 3 meters
- Acceleration due to gravity (g) = 9.8 m/s²

Using the kinematic equation: v_f² = v_i² + 2 * g * (y - y_0), where v_f = 0 m/s (at the height y)

0 = v_i² + 2 * 9.8 m/s² * (3 m - 14 m)
0 = v_i² + 2 * 9.8 m/s² * (-11 m)
v_i² = 2 * 9.8 m/s² * 11 m
v_i² ≈ 215.6 m²/s²
v_i ≈ 14.7 m/s (rounded to one decimal place)

Thus, the correct answer for part C is approximately 14.7 m/s.

In summary, the corrected answers are:
A. Approximately 0.6 seconds
B. Approximately 15.764 meters
C. Approximately 14.7 m/s