please solve this question (a ball is thrown up with velocity u it comes back to the ground with velocity v .take air resistance F has constant if time of ascent is t1 and time of decent is t2 . then find the ratio of t2/t1 and v/u
Answer (1)
7th Apr, 2020
Hello,
Air resistance is constant and same in both cases, so I am ignoring that.
So, we threw a ball with velocity 'u'.
According to formula: v=u+at, we get 0=u+ ( -10 ) ( t1 ) ---> u=10 (t1)
This is because,
- the final velocity is zero, as at max height, it will stop and then come down.
- 10 is the acceleration due to gravity, rounded off.
- t1, as given is time of ascent.
For case 2, v= 0 + 10 (t2) ----> v=10 (t2).
Now , finding ratios by dividing the equations obtained.
t2/t1 = v/u. This is the required answer.
In reality, t1=t2. But since values are not given in the question, we assume them to be different.
Hope this helps.
Thanks.
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