Physics Projectile Motion Help

A 0.36 kg rock is projected from the edge of
the top of a building with an initial velocity of
8.36 m/s at an angle 45◦
above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 13.8 m from the base
of the building
How tall is the building? Assume the
ground is level and that the side of the building is vertical. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m

Horizontal component of launch = cos 45 x 8.36, = 5.91m/sec.
Initial vertical component will be the same.
Time in air = (13.8/5.91) = 2.335 secs.
Time rock is rising = (v/g) = 5.91/9.8, = 0.6031 sec.
Height rising = (v^2/2g) = 1.782m.
Time to fall to ground = (2.335 - 0.6031) = 1.7319 secs.
Height of fall = 1/2 (t^2 x g) = 14.69744m.
Height of building = (14.69744 - 1.782) = 12.915 metres.

Okay so there are two parts of this question. But in part one and two we are trying to figure out the TIME that the flight takes. If we know the amount of time it took for the ball to fall from the top to the bottom then we can just use D = at^2/2 to find D which would be the height of the building.

A.
First find the velocities fo the projectile in the x and y direction use trigonometry. Almost always do this.
-We are given V = 8.36 so

Vx = cos 45 * V = 5.9114 m/s

Vy = sin 45 * V = 5.9114 m/s since the angle was 45 degress the x and y velocities turn out the same.
-
-Now the y- velocity of the projectile is going to be the same as the velocity of the when it went up:
-It will go up at a speed at 5.9114 m/s and decrease because of gravity until it reaches zero then gravity will accelerate it back down. When it reaches back to the top of the building the Vy will be 5.9114 m/s.

-Lets keep track of TIME! Use the equation V = a t. We ahve the velocity of the ball when it reaches back down and we have accelation so lets find the time it would take for the ball to go from its peak back down to the top of the building.