Question 33247
Use this link to see a sketch of the rocket travelling over the building.
Copy the imageshack url into the address window of your browser and click on GO.

http://img146.imageshack.us/my.php?image=rocketbuilding8nu.jpg 

Vx=Vcos@,  Vy=Vsin@
using the equation of motion, s= ut+0.5atē
Sy=Vy*t - 0.5gtē -----------------(1)
When the rocket reaches the pond, Sy=0, and t = T where T is the time of total flight
0=T(Vy - 0.5gT)
Vy = 0.5gT
T = 2Vy/g --------------------------(2)
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Sx = Vx*T -------------------------(3)
Sx = Vx*2Vy/g
120 = 2Vēcos@sin@/g
Vēsin(2@) = 120g --------------(4)
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let t be the time when the rocket reaches 5m above the building. From the geometry of the flight. we know Sx = 100m and Sy = 35m.
Using (3),
100 = Vx*t
t = 100/Vx ------------------------(5)
Using (1),
35 = Vy*t - 4.9tē
35 = Vy*(100/Vx) - 4.9*100ē/Vēx
0.35 = Vy/Vx - 490/Vēx
0.35 = Vsin@)/Vcos@ - 490/Vēx
0.35 = tan@ - 490/Vēcosē@
substituting for Vē from (4),
490/vēcosē@ = tan@ - 0.35
490/(tan@ - 0.35) = Vēcosē@ = (120g/sin(2@))*cosē@
490/(tan@ - 0.35) = 120g/(2sin@cos@)*cosē@
490/(tan@ - 0.35) = 60g/tan@
490tan@ = 60gtan@ - 0.35*60g
490tan@ - 588tan@ = -205.8
90tan@ = 205.8
tan@ = 205.8/98 = 2.1
@ = 64.537 deg
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Using (4),
Vēsin(129.074) = 120g = 1176
Vē = 1514.8
V = 38.92 m/s
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The equation for the parabola formed by the trajectory is, using (1)

Sy=Vy*t - 0.5gtē
Sy = Vsin@*t - 4.9tē
Sy = 38.92*sin(64.537)*t - 4.9tē
Sy = 35.14*t - 4.9tē
==============
The green line is 35 m above the ground.
{{{ graph( 300, 200, 0, 8, 0, 80, 35.14*x - 4.9*x^2,35) }}}