Question

Two bodies are thrown from the same
point with the same velocity of 50ms!. If
their angles of projection are
complimentary to each other and the
difference of maximum heights is 30m,
the minimum and maximum heights are
(g = 10 m/s)​

Answer 1

Suppose the angle of projection of one of them is θ, then the angle of projection of another one is 90−θ. Now if the projection velocity is v then the maximum height is given by

h1=v2sin2θ2g

and

h2=v2sin2(90−θ)2g=v2cos2θ2g

So according the given problem we have

v2sin2θ2g−v2cos2θ2g=30⇒v22g(sin2θ−cos2θ)=30⇒cos(2θ)=30×2gv2=30×2×10502=0.24⇒2θ=76.11°⇒θ=38.05°

So the heights are

h1=v2sin2θ2g=502sin2(38.05)2×10=47.5 m

and

h1=v2cos2θ2g=502×cos2(38.05)2×10=77.5