To find the parachutist's velocity upon reaching the ground, we need to analyze two phases of his descent: free fall and parachute descent.
Phase 1: Free Fall
During the first 10 seconds of free fall, we can calculate the distance fallen and the velocity just before the parachute opens.
- Initial velocity (u) = 0 m/s
- Acceleration (a) = g = 10 m/s²
- Time (t) = 10 s
Using the formula for distance:
d = ut + (1/2)at²
Substituting the values:
d = 0 + (1/2) × 10 × (10)² = 500 m
Now, to find the final velocity (v) after 10 seconds:
v = u + at
v = 0 + 10 × 10 = 100 m/s
Phase 2: Parachute Descent
After 10 seconds, the parachute opens, and the parachutist descends with a net retardation of 2.5 m/s². The effective acceleration during this phase is:
a = g - retardation = 10 - 2.5 = 7.5 m/s²
Now, we need to determine the time taken to reach the ground from the height remaining after the first phase:
Initial height = 2495 m
Height fallen during free fall = 500 m
Remaining height = 2495 - 500 = 1995 m
Using the equation of motion to find the time (t) taken to fall 1995 m:
d = vt + (1/2)at²
Here, initial velocity (v) = 100 m/s, distance (d) = 1995 m, and acceleration (a) = 7.5 m/s².
Rearranging gives:
1995 = 100t + (1/2)(7.5)t²
This simplifies to:
3.75t² + 100t - 1995 = 0
Using the quadratic formula, we can solve for t:
t = [-b ± √(b² - 4ac)] / 2a
Where a = 3.75, b = 100, and c = -1995.
Calculating the discriminant:
b² - 4ac = 10000 + 29925 = 39925
Now, substituting into the formula:
t = [-100 ± √39925] / (2 × 3.75)
Calculating gives us a positive time value.
Finally, we can find the final velocity just before hitting the ground using:
v = u + at
Where u is the velocity at the start of the parachute descent (100 m/s) and a is the effective acceleration (7.5 m/s²).
After calculating, we find that the final velocity upon reaching the ground is:
15 m/s
Final Answer
The parachutist's velocity on reaching the ground will be 15 m/s, which corresponds to option (C).
