
A balloon is rising vertically upwards at a velocity of $10 \mathrm{m} / \mathrm{s} .$ When it is at a height of $45 \mathrm{m}$ from the ground, a parachutist bails out from it. After 3 seconds he opens this parachute and decelerates at a constant rate of $5 \mathrm{m} / \mathrm{s}^{-2} .$ What was the height of the parachutist above the ground when he opened his parachute?(take $g=10 \mathrm{m} / \mathrm{s}^{-2}$ )
(A) 15 m
(B) 30 m
(C) 45 m
(D) 60 m
Answer
198.9k+ views
Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.
Complete step by step answer
We should know that Newton's second law is often stated as \[F=ma\], which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a). This means the more mass an object has, the more force you need to accelerate it. And the greater the force, the greater the object's acceleration.
It is given that:
$\mathrm{u}=10 \mathrm{m} / \mathrm{s}$ in upward direction.
$\mathrm{g}=-10 \mathrm{m} / \mathrm{s}^{2}$ in downward direction
$\mathrm{t}=3 \mathrm{sec}$
$\mathrm{H}=45 \mathrm{m}$
So, now by second equation of motion we get:
$\mathrm{s}=\mathrm{ut}+\dfrac{1}{2} \mathrm{at}^{2}$
Now we have to put the values in the above expression to get the following expression:
$=10 \times 3+\dfrac{1}{2}(-10) \times 3 \times 3$
$=30-45$
$=-15 \mathrm{m}$
Here negative sign indicates that it is directed downwards.
So, height from the ground when he opened parachute is \[45-15=30m\]
So, the height of the parachutist above the ground is 30 m.
Hence the correct answer is option B.
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Complete step by step answer
We should know that Newton's second law is often stated as \[F=ma\], which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a). This means the more mass an object has, the more force you need to accelerate it. And the greater the force, the greater the object's acceleration.
It is given that:
$\mathrm{u}=10 \mathrm{m} / \mathrm{s}$ in upward direction.
$\mathrm{g}=-10 \mathrm{m} / \mathrm{s}^{2}$ in downward direction
$\mathrm{t}=3 \mathrm{sec}$
$\mathrm{H}=45 \mathrm{m}$
So, now by second equation of motion we get:
$\mathrm{s}=\mathrm{ut}+\dfrac{1}{2} \mathrm{at}^{2}$
Now we have to put the values in the above expression to get the following expression:
$=10 \times 3+\dfrac{1}{2}(-10) \times 3 \times 3$
$=30-45$
$=-15 \mathrm{m}$
Here negative sign indicates that it is directed downwards.
So, height from the ground when he opened parachute is \[45-15=30m\]
So, the height of the parachutist above the ground is 30 m.
Hence the correct answer is option B.
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Equation of Trajectory in Projectile Motion: Derivation & Proof

Atomic Structure: Definition, Models, and Examples

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Motion In A Plane Class 11 Physics Chapter 3 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units and Measurements - 2025-26

NCERT Solutions for Class 11 Physics Chapter 3 Motion In A Plane 2025-26
