When a balloon rising vertically upwards at a velocity of 10 m per second is at a height of 45 m from the ground a parachutist bails out from the balloon after 3 second he opens the parachute and deaccelerates at a constant rate of 5 metre per second square then find

1. what is the height of the parachutist above the ground when he opens the parachute.

2. how far is he from the balloon at this instant.

3. with what velocity does he strike the ground.

4. what time does he take in striking the ground after his exit from the balloon.
(take g=10 metre per second square)

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Solution

Step 1: Given
Initial velocity(u)=10 m/s (in upward direction)
Acceleration due to gravity(g)=-10 m/s² (in downward direction)
Acceleration of parachute(a)= 5 m/s²
Time(t)=3 sec
Let V, T, T', and V' are final velocity after 3 sec, time to reach on the ground by parachute, the total time to reach the ground and the velocity with which man reaches the ground.
Step 2: Formula used
$s = u t + \frac{1}{2} a t^{2}$ (second equation of motion)
V=u+gt (first equation of motion)
Step 3: Finding the height from the ground when he opened parachute
$s = u t + \frac{1}{2} a t^{2}$
$= 10 \times 3 + \frac{1}{2} \times (- 10) 3^{2}$
=30-45 s
=-15 m
Here negative sign indicates that it is directed downwards.
So, the height from the ground when he opened parachute is =45-15=30 m
Step 4: Finding the distance from the balloon when he open parachute
time(t) = 3sec
balloon from the ground= 30m
balloon velocity in upward direction=10 m/s
total distance covered by balloon in upward direction= 10*3= 30 m(in upward)
total distance covered by man(parachute) in downward direction= 15 m(in downward)
Hence parachute is 30+15=45 m away from the balloon.
Step 5: Finding the velocity of parahute
(we will take upward direction as positive and downward direction as negative)
time=3sec
V=u+gt
=10 -10 x 3 (gravity act in downward direction)
=-20 m/s (directed downwards)
at time t=3 sec
u=-20 m/s
Remaining displacement after 3 sec(s₁)=-30m
a = +5 m/s² upwards
Let time taken by parachute to reach on ground = T
$s = u t + \frac{1}{2} a t^{2}$
$- 30 = 20 \times T + \frac{5}{2} T^{2}$
T²-8T+12=0
T²-6T-2T+12=0
T(T-6)-2(T-6)=0
T = 6 sec or 2 sec
if T = 6 sec
Step 6: Finding the velocity with which he hits the ground will be
V=u+at
V'=-20+5 x 6 =10 m/s
which is in upward direction not possible.
Hence T=2 sec
V'=-20+5 x 2 =-10 m/s which will be in a downward direction.

Step 7: Finding the total time parachutist takes to hit ground
T' = 3 s+2 s = 5 sec
Hence, the total time taken by man to reach the ground is 5 sec.

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