AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two balloons are simultaneously released from two buildings A and B. Balloon from A rises with constant velocity 10 ms^-1, While the other one rises with constant velocity of 20 ms^-1 . Due to wind the balloons gather horizontal velocity V_x = 0.5 y, where y is the height from the point of release. The buildings are at a distance of 250 m and after some time t the balloons collide. Choose the correct option(s).

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

difference in height of buildings is 50 m

t = 5 sec

difference in height of buildings is 100 m

t = 10 sec

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$(y_{B} - y_{A}) = 20 t - 10 t = 10 t$ ……..(i)

Difference in the height of building

$(X_{B} - X_{A}) = \int_{0}^{t} (0 .5 y_{B}) dt - \int_{0}^{t} (0 .5 y_{A}) dt$ …….(ii)

Separation between the buildings = 250 m

Hence $\int_{0}^{t} 0 .5 \times 20 t . dt - \int_{0}^{t} 0 .5 \times 10 t . dt = 250$

$5 \int_{0}^{t} t . dt = 250$

Hence $5 \frac{t^{2}}{2} = 250 \Rightarrow t = 10 s$

Difference in the height of building = 10t $= 10 \times 10 = 100 m$

Difference in heights $= y_{2} - y_{1} = 20 t - 10 t = 100 m$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!