
Answer:
Time of flight Ā A is greatest
Explanation:
Let uā , uā, uā be their initial velocity and Īøā , Īøā and Īøā be their angle of projection. They all achieve a common highest height of H.
So
H = uā² sin²θā /2g
H = uā² sin²θā /2g
H = uā² sin²θā /2g
On the basis of these equation we can write
uā sinĪøā =uā sinĪøā=uā sinĪøā
For maximum range we can write
D = uā² sin2Īøā /g
1.5 D = uā² sin2Īøā / g
2 D =uā² sin2Īøā / g
1.5 D / D = uā² sin2Īøā /uā² sin2Īøā
1.5 = uā cosĪøā /uā cosĪøā Ā Ā Ā ( since , uā sinĪøā =uā sinĪøā )
uā cosĪøā >uā cosĪøā
uā sinĪøā < uā sinĪøā
2uā sinĪøā / g < 2uā sinĪøā /g
Time of flight B < Time of flight Ā A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight Ā A is greatest .