Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
105 views
in Physics by (25.1k points)
A balloon starts rising from the ground with an acceleration of `1.25ms^-2`. After 8 seconds, a stone is released from the balloon. After releasing, the stone will:
A. cover a distance of 40 m
B. have a displacement of 50 m
C. reach the ground in 4s
D. begin to move down after being released

Please log in or register to answer this question.

3 Answers

0 votes
by (25.1k points)
Correct Answer - C
0 votes
by (75 points)
Cover a distance of 40 m

a=1.25

T=8

u=0

S=ut+1/at²

S=1/2×1.25×64

S=40m
0 votes
by (49.1k points)

Correct option is C. reach the ground in 4s

Displacement of the balloon, s = ut + \(\frac 12\)at2

The velocity of the balloon, v = u + at

Where,

s = displacement (m)

u = initial velocity (m/s)

v =the final velocity (m/s)

a =the acceleration of the body (ms−2)

t = the time is taken (s)

We are Given with, a balloon starts rising from ground therefore initial velocity u=0 m/s

As it moves upwards its acceleration will be 1.25ms−2 and it reaches a certain height then a stone falls from the balloon after 8s.

i.e. u = 0 m/s,

a = 1.25ms−2

Now we will calculate the distance of the stone above the ground about which it begins to fall from the balloon.

Here, Let s =  h, u = 0m/s, and a = 1.25ms−2, t = 8s

Then, substituting the values in the formula s = ut + \(\frac 12\)at2 we get,

h = 0 + \(\frac 12\)(1.25)82

Therefore, h = 40m.

Stone covers a distance of 40m before reaching the ground.

Next, we will find out the velocity of stone at height 40m,

The velocity of the balloon at this height can be obtained from the formula 

v = u + at

Here, u = 0, a = 1.25ms−2, t = 8s

Then we get,

v = 0 + (1.25)8 = 10ms−1

This velocity becomes the initial velocity (u’) of the stone as the stone falls from the balloon from height h.

Therefore we have now, the initial velocity at height h, u’= 10ms-1

Now let us calculate the total time taken by the stone to reach the ground.

Total motion of the stone, is given by an equation,

 h = \(\frac 12\)gt2 − u′t

Here, h = 40m, u’ = 10ms−1 and t is the time taken by the stone to reach the ground.

Substituting values in kinematic equation, we get

−40 = 10t − \(\frac 12\) × 10t2

After solving,

−40 = 10t − 5t2

5t2 − 10t − 40 = 0

Divide the above equation by 5, we get

t− 2t − 8 = 0

After factorization,

t2 − 4t + 2t − 8 = 0

t(t − 4) + 2(t − 4) = 0

Therefore t - 4 = 0 or t + 2 = 0

Then the value of t comes to be, t = 4 or -2

Ignoring the negative value of the time we get, t = 4s.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...