When her parachute is open,the parachutist continues to fall vertically. Immediately after her parachute opens, she decelerates at 12ms^-2 for 2 seconds before reaching a constant speed and she reaches the ground with this speed. The total time taken by the parachutist to fall the 550m from the helicopter to the ground is T seconds. (b) Sketch a speed-time graph for the motion of the parachutist for 0leqslant tleqslant T (c) Find, to the nearest whole number,the value of T. 1. At time t=0 , a parachutist falls vertically from rest from a helicopter which is hovering at a height of 550m above horizontal ground. The parachutist ,who is modelled as a particle, falls for 3 seconds before her parachute opens. While she is falling,and before her parachute opens, she is modelled as falling freely under gravity. The acceleration due to gravity is modelled as being 10ms^-2 (a) Using this model, find the speed of the parachutist at the instant her parachute opens. (1) (2) (5) In a refinement of the model of the motion of the parachutist, the effect of air resistance is included before her parachute opens and this refined model is now used to find a new value of T. (d) How would this new value of T compare with the value found, using the initial model, in part (c)? (e) Suggest one further refinement to the model , apart from air resistance, to make the model more realistic. (1)

# Explanation:<br /><br />## Step 1:<br />We need to find the speed of the parachutist at the instant her parachute opens. The parachutist falls freely under gravity for 3 seconds. The acceleration due to gravity is \(10 \, \mathrm{m/s^2}\).<br /><br />## Step 2:<br />Use the formula for speed under constant acceleration:<br />### \(v = u + at\)<br />where \(u\) is the initial speed (0 m/s), \(a\) is the acceleration (10 m/s²), and \(t\) is the time (3 s).<br /><br />## Step 3:<br />Substitute the values into the formula:<br />### \(v = 0 + (10 \, \mathrm{m/s^2}) \cdot (3 \, \mathrm{s})\)<br /><br />## Step 4:<br />Calculate the speed:<br />### \(v = 30 \, \mathrm{m/s}\)<br /><br /># Answer: (a) \(30 \, \mathrm{m/s}\)<br /><br /># Explanation:<br /><br />## Step 1:<br />We need to sketch a speed-time graph for the motion of the parachutist for \(0 \leq t \leq T\).<br /><br />## Step 2:<br />From \(t = 0\) to \(t = 3\) seconds, the parachutist accelerates from 0 to 30 m/s under gravity.<br /><br />## Step 3:<br />From \(t = 3\) to \(t = 5\) seconds, the parachutist decelerates at \(12 \, \mathrm{m/s^2}\). The deceleration can be calculated using:<br />### \(v = u + at\)<br />where \(u = 30 \, \mathrm{m/s}\), \(a = -12 \, \mathrm{m/s^2}\), and \(t = 2 \, \mathrm{s}\).<br /><br />## Step 4:<br />Substitute the values into the formula:<br />### \(v = 30 \, \mathrm{m/s} + (-12 \, \mathrm{m/s^2}) \cdot (2 \, \mathrm{s})\)<br /><br />## Step 5:<br />Calculate the speed:<br />### \(v = 30 \, \mathrm{m/s} - 24 \, \mathrm{m/s} = 6 \, \mathrm{m/s}\)<br /><br />## Step 6:<br />From \(t = 5\) seconds onwards, the parachutist falls at a constant speed of 6 m/s until she reaches the ground.<br /><br /># Answer: (b) The speed-time graph will show a linear increase from 0 to 30 m/s in the first 3 seconds, a linear decrease from 30 m/s to 6 m/s from 3 to 5 seconds, and then a horizontal line at 6 m/s from 5 seconds onwards.<br /><br /># Explanation:<br /><br />## Step 1:<br />We need to find the total time \(T\) taken by the parachutist to fall the \(550 \, \mathrm{m}\) from the helicopter to the ground.<br /><br />## Step 2:<br />Calculate the distance fallen in the first 3 seconds using the formula:<br />### \(s = ut + \frac{1}{2}at^2\)<br />where \(u = 0 \, \mathrm{m/s}\), \(a = 10 \, \mathrm{m/s^2}\), and \(t = 3 \, \mathrm{s}\).<br /><br />## Step 3:<br />Substitute the values into the formula:<br />### \(s = 0 + \frac{1}{2}(10 \, \mathrm{m/s^2})(3 \, \mathrm{s})^2\)<br /><br />## Step 4:<br />Calculate the distance:<br />### \(s = \frac{1}{2} \cdot 10 \cdot 9 = 45 \, \mathrm{m}\)<br /><br />## Step 5:<br />Calculate the distance fallen during deceleration (3 to 5 seconds). Use the formula:<br />### \(s = ut + \frac{1}{2}at^2\)<br />where \(u = 30 \, \mathrm{m/s}\), \(a = -12 \, \mathrm{m/s^2}\), and \(t = 2 \, \mathrm{s}\).<br /><br />## Step 6:<br />Substitute the values into the formula:<br />### \(s = (30 \, \mathrm{m/s})(2 \, \mathrm{s}) + \frac{1}{2}(-12 \, \mathrm{m/s^2})(2 \, \mathrm{s})^2\)<br /><br />## Step 7:<br />Calculate the distance:<br />### \(s = 60 - 24 = 36 \, \mathrm{m}\)<br /><br />## Step 8:<br />Calculate the remaining distance to be covered at constant speed:<br />### \(\text{Remaining distance} = 550 \, \mathrm{m} - 45 \, \mathrm{m} - 36 \, \mathrm{m} = 469 \, \mathrm{m}\)<br /><br />## Step 9:<br />Calculate the time taken to cover the remaining distance at constant speed (6 m/s):<br />### \(t = \frac{\text{distance}}{\text{speed}} = \frac{469 \, \mathrm{m}}{6 \, \mathrm{m/s}}\)<br /><br />## Step 10:<br />Calculate the time:<br />### \(t \approx 78.17 \, \mathrm{s}\)<br /><br />## Step 11:<br />Add up all the times to find the total time \(T\):<br />### \(T = 3 \, \mathrm{s} + 2 \, \mathrm{s} + 78.17 \, \mathrm{s} \approx 83 \, \mathrm{s}\)<br /><br /># Answer: (c) \(83 \, \mathrm{s}\)<br /><br /># Explanation:<br /><br />## Step 1:<br />We need to compare the new value of \(T\) with the value found using the initial model in part (c).<br /><br />## Step 2:<br />Including air resistance before the parachute opens will slow down the parachutist's fall, resulting in a longer time to reach the ground.<br /><br /># Answer: (d) The new value of \(T\) will be greater than the value found in part (c).<br /><br /># Explanation:<br /><br />## Step 1:<br />Suggest one further refinement to the model to make it more realistic.<br /><br />## Step 2:<br />Consider the variation in gravitational acceleration with altitude, as gravity decreases slightly with height above the Earth's surface.<br /><br /># Answer: (e) Consider the variation in gravitational acceleration with altitude.