Question

A body is to be projected at an angle with a velocity 24m/s just to pass over an obstacle 14m high at adistance of24m.then angle is what.

Answer 1

Do the vertical problem first
Vi = initial speed up
then
v = Vi - g t
at top, v = 0
0 = Vi - g t
Vi = g t
t = Vi/g

h = 0 + Vi t - .5 g t^2
14 = Vi^2/g - .5 Vi^2/g
28 g = Vi^2
Vi = sqrt (28 g)
Vi = 16.6 m/s

Vi = v sin theta
16.6 = 24 sin theta
so ....