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: v
The 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]
Answer:
v = sqrt (2ay)
Step-by-step explanation:
I just took the test, and this is the correct answer!!!
