how you do this problem?

Question

A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, forming a 1.00 m high railing around the playground. A ball falls to the street below, and a passerby returns it by launching it at an angle of 53.0° above the horizontal at a point 25.0 m from the base of the building wall. The ball takes 2.00 s to reach a point vertically above the wall.
(a) Find the speed at which the ball was launched.
m/s
(b) Find the vertical distance by which the ball clears the wall.
m
(c) Find the distance from the wall to the point on the roof where the ball lands.
m

Answer 1

x(t)=vi*cos(53)*t

from the statement, we have
25=vi*cos(53)*2
vi=20.77 m/s

The height of the ball is
y(t)=20.77*sin(53)*t-.5*9.81*t^2
y(2)=13.56 m
so it clears the wall by 6.56 m

when y(t)=6, the ball is at the same height as the roof.
one solution is 0.412 s, but we know that is before the ball clears the wall
the other is
2.97 seconds
x(2.97)=37.125
the distance from the railing is
37.125-25
=12.125 m

j

Answer 2

Set up the origin of the coordinates in x,y to the time t=0
your acceleration is
ax=0
ay=-g

after integration of a you get v(t)

vx(t)=v0*cos(alpha)
vy(t)=v0*sin(alpha) - g*t

the second integration gives you the way r(t)

x(t)=t*v0*cos(alpha)
y(t)=t*v0*sin(alpha) - 0.5g*t^2

a) let be x=xp=25 m and t=tp=2 s
To the time tp has the ball reached a vertical position over the wall
x(t=tp=2s)=25 m=tp*v0*cos(alpha)
v0=xp/[tp*cos(alpha)]
v0=20.77 m/s

b) to the same time tp has the ball the position yp +7 m over the playground.
y(t=tp=2s)=yp+7m=tp*v0*sin(alpha) - 0.5g*(tp)^2
yp=2s-(20.77 m/s)*sin(53 degrees) - 0.5*(9.81 m/s^2)*4s^2 -7m
yp=6.56 m

c) y(t)=tp*v0*sin(alpha) -0.5g*t^2
The track of the ball follows a parabola. The quadratic equation will deliver you 2 solutions:
t1 = time when the ball is landing in distance xL+25 m from origin
t2 = time when the ball reached the same height (y=6m) before passing the wall.
Note that the ascending time and descending time are that identical.
with
t^2 + pt + q = 0
p=-(2/g)*v0*cos(alpha)
q=(2*6m)/g =12m/g
you get
t1=2.97 s
t2=0.412 s
x(t=tL=t1=2.97s) = xL+25 m = vo*t1*cos(alpha)
xL=12.13 m