Question

A parachutist jumps out of an airplane and accelerates with gravity for 6 seconds. He then pulls the parachute cord and after a 4 s deceleration period, descends at 10 m/s for 60 seconds, reaching the ground. From what height did the parachutist jump? Assume acceleration due to gravity to be 10 m/s² throughout the motion.

• A. 840 m
• B. 920 m
• C. 980 m
• D. 1020 m

Answer 1

Answer: (B)

920 m

Answer 2

Let the motion of the parachutist be divided into three phases:

Phase 1: Free fall under gravity. The parachutist jumps out of the airplane, so the initial velocity is $u_1 = 0$. The motion lasts for $t_1 = 6$ seconds. The acceleration is due to gravity, $a_1 = g = 10 \, m/s^2$ (downwards). The distance covered in this phase is $h_1$. Using the kinematic equation $h = ut + \frac{1}{2}at^2$: $h_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2$ $h_1 = (0)(6) + \frac{1}{2} (10)(6)^2$ $h_1 = 0 + \frac{1}{2} (10)(36) = 5 \times 36 = 180 \, m$. The velocity at the end of this phase, $v_1$, is given by $v = u + at$: $v_1 = u_1 + a_1 t_1 = 0 + (10)(6) = 60 \, m/s$ (downwards).

Phase 2: Deceleration period after pulling the parachute cord. This phase starts with the velocity $u_2 = v_1 = 60 \, m/s$. The duration of this phase is $t_2 = 4$ seconds. After this deceleration period, the parachutist descends at a constant velocity of 10 m/s. This constant velocity is the final velocity of Phase 2, so $v_2 = 10 \, m/s$. The distance covered in this phase is $h_2$. We can find $h_2$ using the average velocity since the acceleration is constant (implicitly, as it's a fixed deceleration period): $h_2 = \left(\frac{u_2 + v_2}{2}\right) t_2$ $h_2 = \left(\frac{60 + 10}{2}\right) \times 4$ $h_2 = \left(\frac{70}{2}\right) \times 4 = 35 \times 4 = 140 \, m$. Alternatively, we could find the acceleration $a_2$ first: $v_2 = u_2 + a_2 t_2 \Rightarrow 10 = 60 + a_2(4) \Rightarrow 4a_2 = -50 \Rightarrow a_2 = -12.5 \, m/s^2$. Then, $h_2 = u_2 t_2 + \frac{1}{2} a_2 t_2^2 = (60)(4) + \frac{1}{2}(-12.5)(4)^2 = 240 - 6.25 \times 16 = 240 - 100 = 140 \, m$.

Phase 3: Descent at constant velocity. This phase starts with the velocity $u_3 = v_2 = 10 \, m/s$. The velocity is constant, so $a_3 = 0$. The duration of this phase is $t_3 = 60$ seconds. The distance covered in this phase is $h_3$. Since the velocity is constant, distance = velocity × time: $h_3 = u_3 t_3 = (10 \, m/s)(60 \, s) = 600 \, m$.

The total height from which the parachutist jumped is the sum of the distances covered in each phase: Total height $H = h_1 + h_2 + h_3$ $H = 180 \, m + 140 \, m + 600 \, m$ $H = 320 \, m + 600 \, m = 920 \, m$.