The formula gives the maximum height y of a projectile launched straight up, given acceleration a and initial velocity v. y=v22a Solve for v. v=2ay‾‾‾‾√ v=4y2a2 v=4a2y2 v=2ay√a

Answer:Hence, the expression is:[tex]v=\sqrt{2ay}[/tex]Step-by-step explanation:A projectile is an object that is given an initial velocity and is acted on by gravity.The maximum height attained by the object is the highest vertical position along it's trajectory.Let acceleration is given by: a and initial velocity is given by: vThe maximum height(y) is given by:[tex]y=\dfrac{v^2\times \sin \theta}{2\times a}[/tex]where thetha( θ) is the launch angle.As the projectile is launches straight up.Hence, θ=90°.Hence,sin θ=1.Hence,[tex]y=\dfrac{v^2}{2a}\\\\v^2=2ay\\\\v=\sqrt{2ay}[/tex]Hence, the expression is:[tex]v=\sqrt{2ay}[/tex]