u=o 7.A helicopter is hovering at rest above horizontal ground at the point H.A parachutist steps out of the helicopter and immediately falls vertically and freely under gravity from rest for 2.5s.His parachute then opens and causes him to immediately decelerate at a constant rate of 3.9ms^-2 for T seconds (Tlt 6) , until his speed is reduced to Vms^-1 . He then moves with this constant speedVms^-1 until he hits the ground. While he is decelerating he falls a distance of 73.75 m. The total time between the instant when he leaves H and the instant when he hits the ground is 20 s. The parachutist is modelled as a particle. (a) Find the speed of the parachutist at the instant when his parachute opens. (1) (b) Sketch a speed-time graph for the motion of the parachutist from the instant when he leaves H to the instant when he hits the ground. (2) (c) Find the value of T. (5) (d) Find, to the nearest metre,the height of the point H above the ground. 1 b

(a) To find the speed of the parachutist at the instant when his parachute opens, we can use the equation of motion under constant acceleration: where:- is the final velocity,- is the initial velocity (which is 0 since he starts from rest),- is the acceleration (which is due to gravity, and it's negative because it's acting downwards),- is the time.Given that the parachutist falls for 2.5 seconds before the parachute opens, we can substitute these values into the equation: So, the speed of the parachutist at the instant when his parachute opens is .(b) To sketch a speed-time graph for the motion of the parachutist, we need to consider the different phases of his motion:1. From the instant he leaves H to the instant when his parachute opens, he is accelerating downwards due to gravity. This will be represented by a straight line with a negative slope of .2. From the instant his parachute opens to the point where his speed is reduced to , he is decelerating at a constant rate of . This will be represented by a straight line with a negative slope of .3. From the point where his speed is reduced to to the instant when he hits the ground, he is moving with a constant speed of . This will be represented by a horizontal line.(c) To find the value of , we can use the equation of motion under constant acceleration: where:- is the final velocity,- is the initial velocity (which is since he is decelerating),- is the acceleration (which is due to the deceleration),- is the time.Given that the parachutist decelerates from to in time , we can rearrange the equation to solve for : Substituting the given values: (d) To find the height of point H above the ground, we need to consider the different phases of the parachutist's motion:1. From the instant he leaves H to the instant when his parachute opens, he falls freely under gravity. We can use the equation of motion: where:- is the distance fallen,- is the initial velocity (which is 0 since he starts from rest),- is the acceleration (which is due to gravity),- is the time.Given that he falls for 2.5 seconds before the parachute opens, we can substitute these values into the equation: So, the height of point H above the ground is .