A body is projected with velocity u so that the maximum height is thrice the horizontal range. Then the maximum height is??
Answers
Answered by
13
max height=u^2 sin^2 Q/2g......(i)
range= u^2 sin2Q/g........(ii)
so, put equate them and put value of g=10m/s^2
your sinQ will come 12/root 145
hence put this in eq. (i) ur max height will come u^2*144?g*145
range= u^2 sin2Q/g........(ii)
so, put equate them and put value of g=10m/s^2
your sinQ will come 12/root 145
hence put this in eq. (i) ur max height will come u^2*144?g*145
Answered by
5
Answer: 72u²/145g is the answer dude
Explanation is in attachment
Explanation:
Attachments:
Similar questions