Please wait...
Contact Us
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry



Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

9321924448 / 9987178554

Mon to Sat - 10 AM to 7 PM

Solve this

Asked by venkat 25th December 2017, 8:56 PM
Answered by Expert
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!

Chat with us on WhatsApp