A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3m/s² for 0.5 minute. If the maximum height reached by it is 80m then the angle of projection is[g=10m/s²]

Answer 1

Vh = 3 m/s² * 30 s = 90 m/s

½ m Vv² = m g h
... Vv = √(2 g h) = √( 2 * 10 * 80)

tan(Θ) = Vv / Vh

Answer 2

v=u+at
v=0+3×30 (0.5×60=30)
v=90
u cosx=90......... eq (1)
H=u^2sin^2x÷2g=80 (given)
u sinx=40 (solving)......eq (2)
eq (1)÷eq (2)
tanx=4/9
x=tan^-1 (4/9)