a slingshot launches a rock straight upward at 25 m/s, how high does the rock go, how long is it in the air, what is the velocity and the acceleration of the rock at its highest point

g = 9.81 m/s^2

v = Vi - g t
at top v = 0
t = 25/9.81 seconds to top
double that for time in air

m g h = (1/2) m Vi^2
h = 625/(2*9.81)

v is 0 at top of course
a is -9.81 m/s^2 forever until it hits the mud.

To solve this problem, we can use some basic equations of motion. Let's break it down step by step:

1. Finding the time taken to reach the highest point:
We know the initial vertical velocity of the rock is 25 m/s, and the acceleration due to gravity (in the opposite direction) is approximately -9.8 m/s². In this case, the rock is moving upward, so the final velocity at the highest point would be zero. We can use the equation:

v = u + at

where:
v = final velocity (zero in this case)
u = initial velocity (25 m/s)
a = acceleration due to gravity (-9.8 m/s²)
t = time

Rearranging the equation, we have:

t = (v - u) / a

Substituting the known values, we get:

t = (0 - 25) / -9.8
t ≈ 2.55 seconds

Therefore, it takes approximately 2.55 seconds for the rock to reach its highest point.

2. Calculating the maximum height reached by the rock:
To find the maximum height, we can use the kinematic equation:

s = ut + (1/2)at²

where:
s = vertical displacement or height
u = initial velocity (25 m/s)
t = time (2.55 seconds)
a = acceleration due to gravity (-9.8 m/s²)

Substituting the known values, we have:

s = (25 × 2.55) + (0.5 × -9.8 × 2.55²)
s ≈ 32 meters

Therefore, the rock reaches a height of approximately 32 meters.

3. Finding the total time of flight:
The total time of flight is the time taken to reach the maximum height and then fall back down. Since the time taken to reach the highest point is 2.55 seconds, the total time of flight will be twice that:

Total time of flight = 2 × t
Total time of flight ≈ 2 × 2.55
Total time of flight ≈ 5.1 seconds

Therefore, the rock is in the air for approximately 5.1 seconds.

4. Determining the velocity and acceleration at the highest point:
At the highest point, the velocity of the rock is zero since it briefly comes to a stop before falling back down. The acceleration due to gravity, however, remains constant at -9.8 m/s² throughout the motion. Therefore, the velocity at the highest point is 0 m/s, and the acceleration at the highest point is -9.8 m/s².

In summary:
- The rock reaches a maximum height of approximately 32 meters.
- It is in the air for approximately 5.1 seconds.
- The velocity of the rock at its highest point is 0 m/s.
- The acceleration of the rock at its highest point is -9.8 m/s².