Question

A ball is thrown vertically upward its velocity is 10 M per second when it reaches half of its maximum height how high did the wall rise

Answer 1

I think the question will be:
A ball is thrown vertically upward with velocity of 10 m/s. How does the ball did rise when it reaches distance of half of its max. height.
v=0
u=10m/s
g=-9.8 m/s^2
s=maximum distance covered
Then,
v^2-u^2=2×g×s
=)0-10×10=2×-9.8×s
=)s=-100/-19.6
=)s=5.1 m
Now,
The distance it covered when it reaches half of its max. height
=5.1/2
=2.55 m
Hope it helps u.

Answer 2

Let the total height be x.

Velocity at x/2 = 10m/s

Assume u=10,v= 0, g= 10(approx).

And s be half distance above x/2.

Then v^2 = u^2 - 2*g*s(for upward motion)

=) 0^2 = 10^2 - 2*10*s

=) 0= 100-20s

=) 20s = 100

=) s= 100/20 = 5m.

Means x-x/2 = 5m

=) x = 10m.

Hope it's helpful to u.