Find solution
Physics calculation- does anyone understand the solution to this question?

Hello..a physics question for all of u..ihave a doubt..can anyone post the solved answer?

  • okk.. A ball thrown up vertically returns to the thrower after 6 secs. Find 1. the veolocity with which the ball was thrown up 2. the maximum height the ball reached 3. its position after 4 secs. i didnt attend school for 3 days, so i was not able to understand this question kindly help me by solving it thanks a lot for your help!

  • Answer:

    If thrown up vertically, the ball will slow down the same way as it is speeding up when it comes down. So it will go up for 3 seconds. v = v° - gt and at the maximum height v = 0 v° = g * t = 9,8 * 3 = 29,4 m/s s = gt²/2 = 9,8*3²/2 = 44,1 m After 4 seconds, the ball has fallen for 1 second. s = gt²/² = 4,9 m So it is at 44,1 - 4,9 = 39,2 m

Xin at Yahoo! Answers Visit the source

Was this solution helpful to you?

Other answers

1) v= u -9.8t, v is 0 at top consider t= 3 seconds (half of 6 s) because the gravity remains same on to and fro motion u= 9.8 X 3 = 29.4 m/s 2) s= ut+0.5 (-9.8)t^2 s= 29.4 X 3 - 0.5 X 9.8 X 9 2. s= 44.1 m 3) s1= 0+ 0.5 X 9.8 X 1 s1 from top = 4.9 m Therefore position of ball after 4 s= 44.1-4.9 = 39.2 m

мbтм™

you should the give the position of the ball in 3secs or the position of the ball in 6 secs

Feenash

Related Q & A:

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.