The correct option is D √α2β
y=βx2
Differentiating w.r.t t,
dydt=2βxdxdt=2βxvx
Differenciating w.r.t t again,
⇒d2ydt2=2βvxdxdt+2xβdvxdt
⇒d2ydt2=2βv2x+2xβdvxdt
dvxdt=0 (∵ There is no acceleration in the horizontal direction
d2ydt2=α (given)⇒α=2βv2x∴vx=√α2β
Alternate solution:
y=βx2
Comparing with the equation of the trajectory,
y=xtanθ−gx22u2cos2θ
tanθ=0⇒θ=0 ...(1)g2u2cos2θ=β ...(2)
acceleration in y direction, g=α ..(3)
Putting (1) and (3) in (2), we get
α2u2(cos20)=β⇒u2=α2β⇒u=√α2β