Solve this
Asked by venkat
25th December 2017,
8:56 PM
Answered by Expert
Answer:
Since Q is projected after 1 second, P will reach the maximum height and meet Q when it is returning back.
Let maximum height reached by P is H.
H = (u×u) /(2×g), where u is initial projection velocity 30 m/s
H = (30×30) / (2*10) = 45 m.
after reaching the maximumheight let P travels back for t seconds and meet Q.
Let the distance travelled by P in that t seconds is h and it is given by
h = (1/2)g×t2 = (1/2)×10×t2 = 5×t2 .........................(1)
Then distance travelled by Q when it meets P is 45-h and the time taken is (2+t)s. This distance is given by
45-h = 30×(2+t) - (1/2)×10×(2+t)2
after substituting for h from eqn.(1), we will get
45 - 5×t2 = 30×(2+t)-5×(2+t)2
solving the above eqn for t, we get t = 0.5 s.
Hence P travelled 3.5 s before meetin Q
Answered by Expert
27th December 2017,
3:13 PM
Rate this answer
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
You have rated this answer 10/10
Your answer has been posted successfully!
RELATED STUDY RESOURCES :
JEE QUESTION ANSWERS
Browse free questions and answers by Chapters
- 1 Thermodynamics
- 2 Gravitation
- 3 Electromagnetic Waves
- 4 Communication Systems
- 5 Laws of Motion
- 6 Current Electricity
- 7 Work, Energy and Power
- 8 Kinematics
- 9 Physics and Measurement
- 10 Rotational Motion
- 11 Properties of Solids and Liquids
- 12 Kinetic Theory of Gases
- 13 Oscillations and Waves
- 14 Electrostatics
- 15 Magnetic Effects of Current and Magnetism
- 16 Electromagnetic Induction and Alternating Currents
- 17 Optics
- 18 Dual Nature of Matter and Radiation
- 19 Atoms and Nuclei
- 20 Electronic Devices
Trending Tags
Latest Questions
JEE Main Mathematics
Asked by amaladelsingh
28th October 2020,
10:24 PM
CBSE X Biology
Asked by reenacharan2008
28th October 2020,
9:11 PM
