Question 5

A parachutist after bailing out falls $$50$$ m without friction. When parachute opens, it decelerates at $$2$$ m/s$$^2$$. He reaches the ground with a speed of $$3$$ m/s. At what height, did he bail out?

Solution

Solution & Explanation

1. Analyze Phase 1: Free Fall (Before Parachute Opens)

The parachutist bails out from rest, meaning his initial velocity ($$u_1$$) is zero:

$$u_1 = 0 \,\, \text{m/s}$$

He falls a distance of $$h_1 = 50 \,\, \text{m}$$ under gravity without any air friction. Taking downward as the positive direction, his acceleration during this phase is $$a_1 = g = 9.8 \,\, \text{m/s}^2$$. We find his velocity ($$v_1$$) just as the parachute opens using the third equation of motion ($$v^2 = u^2 + 2as$$):

$$v_1^2 = u_1^2 + 2 \cdot g \cdot h_1$$

$$v_1^2 = 0^2 + 2 \cdot 9.8 \cdot 50$$

$$v_1^2 = 980 \,\, \text{m}^2/\text{s}^2$$

2. Analyze Phase 2: Deceleration (After Parachute Opens)

When the parachute opens, the system begins to slow down. Let us define the kinematic parameters for this second phase of motion:

Initial velocity for this phase: It matches the final velocity of the free fall, so $$u_2^2 = v_1^2 = 980 \,\, \text{m}^2/\text{s}^2$$.
Uniform deceleration: $$a_2 = -2 \,\, \text{m/s}^2$$ (negative because it opposes downward motion).
Final velocity upon reaching the ground: $$v_2 = 3 \,\, \text{m/s}$$.
Distance traveled during deceleration: Let this be $$h_2$$.

Applying the third equation of motion for this deceleration phase:

$$v_2^2 = u_2^2 + 2 \cdot a_2 \cdot h_2$$

$$(3)^2 = 980 + 2 \cdot (-2) \cdot h_2$$

$$9 = 980 - 4 \cdot h_2$$

Rearranging the equation to isolate the second height segment ($$h_2$$):

$$4 \cdot h_2 = 980 - 9$$

$$4 \cdot h_2 = 971$$

$$h_2 = \frac{971}{4} = 242.75 \,\, \text{m}$$

3. Calculate the Total Bailing Height

The total height ($$H$$) from which the parachutist initially bailed out is the sum of the distance covered during the free fall ($$h_1$$) and the distance covered while decelerating ($$h_2$$):

$$H = h_1 + h_2$$

$$H = 50 + 242.75 = 292.75 \,\, \text{m}$$

Rounding this value to the nearest whole integer gives:

$$H \approx 293 \,\, \text{m}$$

Concept Check: The problem breaks into two distinct acceleration zones. During the initial brief gravity drop ($$50 \,\, \text{m}$$), the jumper builds a massive velocity of nearly $$31.3 \,\, \text{m/s}$$. Because the parachute’s braking force is relatively gentle ($$2 \,\, \text{m/s}^2$$), a long braking runway of over $$242 \,\, \text{m}$$ is required to safely bleed off that velocity down to a walking speed of $$3 \,\, \text{m/s}$$.