Asked by Nabila
A stone is drop from a height 19.6m above the ground while a second stone is simultaneously projected from the ground with sufficient velocity to enable it to accept 19.6m. When and where the stone will meet
Answers
Answered by
bobpursley
dropped stone:
hf=hi+vi*t-4.9t^2
hf=19.6-4.9t^2
thrown stone:
Initial KE =final PE of max height
1/2 mvi^2=mg*19.6
vi=sqrt(2*9.8*19.6)
vi= 19.6m/s
hf=hi+vi*t-4.9t^2
hf=19.6*t-4.9t^2
but the hf is the same for each stone.
19.6*t-4.9t^2 =19.6-4.9t^2
they meet at time t=1second
hf=19.6*t-4.9t^2....
hf=hi+vi*t-4.9t^2
hf=19.6-4.9t^2
thrown stone:
Initial KE =final PE of max height
1/2 mvi^2=mg*19.6
vi=sqrt(2*9.8*19.6)
vi= 19.6m/s
hf=hi+vi*t-4.9t^2
hf=19.6*t-4.9t^2
but the hf is the same for each stone.
19.6*t-4.9t^2 =19.6-4.9t^2
they meet at time t=1second
hf=19.6*t-4.9t^2....
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.