Projectile Motion XY Plane: Solving for Height & Time

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A ball is tossed from a building with an initial velocity of 8.00 m/s at a 20-degree angle below the horizontal, striking the ground after 3.00 seconds. The horizontal distance from the building's base is calculated to be 22.5 meters. To find the height from which the ball was thrown, the equation h = Vx*t + 1/2*g*t^2 is suggested. For determining the time it takes to reach a point 10 meters below the launch level, the same formula can be applied with h set to 10 m. The discussion emphasizes the application of kinematic equations and the need for clarity in solving for time.
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Homework Statement



A ball is tossed from a building. The ball is given an intial velocity of 8.00 m/s at an angle of 20 degrees below the horizontal. It strikes the ground 3.00s later.
a) how far horizontally from the base of the building does the ball strike the ground?
b) Find the height from which the ball was thrown.
c) How long does it take the ball to reach a point 10m below the level of launching

Homework Equations


Kinematic equations.


The Attempt at a Solution


I solved part a with no problem. 22.5m. However, I can't come up with the others using the kinematics or trig. I am missing something somewhere.
The vel in the x direction is 7.5m/s and the vel in the y is 2.74 m/s.
Thats all i got.
 
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For (b) use
h = Vx*t + 1/2*g*t^2
For (c)
Vxf^2 =Vxi^2 + 2*g*h'
 
ok got b). How can I use that equation for c) when it is asking for time?
 
h = Vx*t + 1/2*g*t^2
Use the above formula and put h = 10 m and solve for t.
 
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