The Attempt at a Solution
For part (d),
The curvature radius of trajectory at its
1) initial point = horizontal range/2 =( v02 sin (2α))/2g
2) peak = height of ascent/2 = ( v0 sinα)2 /2g
Is this correct?
In this problem, the time of ascent is equal to the time of descent.
Is there anyway to find it out without calculating the time of ascent and the time of descent ?