CF 102697023 - Ping Pong Parachute

Rating: -
Tags: -
Solve time: 1m 47s
Verified: yes

Solution

Problem Understanding

The problem describes a rocket launch where the only measured value is the initial upward velocity. We need to predict the maximum height reached by the rocket before it starts falling. The rocket reaches its highest point when its velocity becomes zero, so the task is to use the given initial velocity and the constant acceleration caused by gravity to compute the distance traveled upward.

The input contains one floating point value, the initial velocity of the rocket in meters per second. The output should be the maximum height in meters as a floating point number. The relationship between velocity, acceleration, and distance is given by the kinematic equation:

$$V_f^2 = V_i^2 + 2ad$$

At the highest point, $V_f = 0$, and gravity gives $a = -9.8$. Solving for the height gives:

$$d = \frac{V_i^2}{19.6}$$

The input is a single double value, so there is no large data structure or many test cases to process. The algorithm only needs constant time arithmetic. A simulation of the flight would be unnecessary and could introduce precision issues, while the direct formula gives the exact mathematical result expected by the problem.

The main edge cases come from handling floating point values correctly. A rocket with zero starting velocity should produce a height of zero because it never leaves its starting position. For input 0, the correct output is 0.0, while a solution that always assumes upward movement might produce an incorrect positive value.

A very small initial velocity also matters. For input 0.1, the answer is approximately 0.0005102040816326531. Rounding too early or converting the input to an integer would incorrectly output zero.

A larger velocity should still be handled with floating point arithmetic. For input 100, the answer is approximately 510.2040816326531. A careless implementation using integer division could lose the decimal portion.

Approaches

The brute-force approach would be to simulate the rocket's movement over time. We could repeatedly update the velocity by adding the acceleration and add the traveled distance until the velocity reaches zero. This would eventually approximate the maximum height, but the result depends on the chosen time step. A very small time step improves precision while increasing the number of iterations. For a high velocity, thousands or millions of simulation steps might be required, and floating point errors accumulate.

The better approach comes from recognizing that the problem already gives the exact equation describing the motion. The highest point is a specific state where the final velocity is zero. Instead of recreating every moment of the flight, we can directly solve the equation for the unknown height.

The brute-force method works because it follows the physical process, but it fails because the continuous motion has no natural number of simulation steps. The observation that the final velocity is known lets us reduce the entire problem to one formula evaluation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(k), where k is the number of simulation steps	O(1)	Too slow and imprecise
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the initial velocity of the rocket as a floating point number.
Substitute the known values into the motion equation. Since the final velocity at the highest point is zero and acceleration is -9.8, the equation becomes:

$$0 = V_i^2 - 19.6d$$

Rearranging gives:

$$d = \frac{V_i^2}{19.6}$$

Compute the height using the formula and print the result.

The calculation directly models the moment when the rocket stops moving upward. There is no need to track the path between launch and the highest point.

Why it works

The formula comes from the exact relationship between velocity, acceleration, and displacement under constant acceleration. During the ascent, gravity is the only acceleration, so the equation always holds. At the maximum height, the rocket's upward velocity becomes zero. Substituting that condition leaves only one unknown, the height, which the algorithm computes. Since the mathematical model exactly matches the problem description, the returned value is the maximum height.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    v = float(input())
    height = v * v / 19.6
    print(height)

if __name__ == "__main__":
    solve()

The solution reads the velocity as a float because the input is not restricted to integers and the output requires decimal precision.

The multiplication v * v computes the squared initial velocity from the equation. Dividing by 19.6 applies the rearranged kinematic formula. Python floating point arithmetic is sufficient here because the problem only requires the normal printed decimal representation.

There are no boundary indices or integer overflow concerns because the entire computation uses a single floating point value.

Worked Examples

For input:

7.78

the calculation proceeds as follows:

Initial velocity	Squared velocity	Division by 19.6	Height
7.78	60.5284	60.5284 / 19.6	3.0881836734693877

The result matches the sample output. This example confirms that the formula handles ordinary launch speeds correctly.

For input:

the trace is:

Initial velocity	Squared velocity	Division by 19.6	Height
0	0	0 / 19.6	0.0

This demonstrates the case where the rocket has no upward motion and the maximum height remains at the starting point.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a few arithmetic operations are performed
Space	O(1)	Only the input value and the computed height are stored

The solution fits easily within the limits because it does not depend on the size of the input beyond reading one number.

Test Cases

import sys
import io

def solve(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)
    v = float(sys.stdin.readline())
    ans = v * v / 19.6
    out = str(ans)
    sys.stdin = old_stdin
    return out

assert solve("7.78\n") == "3.0881836734693877", "sample 1"
assert solve("0\n") == "0.0", "zero velocity"
assert solve("0.1\n") == "0.0005102040816326531", "small velocity"
assert solve("100\n") == "510.2040816326531", "large velocity"
assert solve("19.6\n") == "19.6", "boundary precision"

Test input	Expected output	What it validates
`7.78`	`3.0881836734693877`	Original sample calculation
`0`	`0.0`	Rocket with no initial movement
`0.1`	`0.0005102040816326531`	Small floating point values
`100`	`510.2040816326531`	Larger values without overflow
`19.6`	`19.6`	Precision of the formula

Edge Cases

For zero velocity, the input is:

The algorithm computes 0 * 0 / 19.6, which gives 0.0. This correctly represents a rocket that never gains height.

For very small velocity, the input is:

0.1

The algorithm keeps the value as a floating point number, computes 0.01 / 19.6, and returns approximately 0.0005102040816326531. An integer conversion would incorrectly remove the entire result.

For a large launch speed, the input is:

The calculation becomes 10000 / 19.6, producing approximately 510.2040816326531. The constant time formula avoids any loop count depending on the rocket's height, so the computation remains accurate and fast.