Question 33247: A toy rocket is launched from a point 60m from a school building. The school building is 40m wide. On the other side of the building is a 15m wide parking lot and next to it is a pond that is 10m wide. The rocket is aimed over the building to land in the center of the pond, 120m away. The building is 30m tall. The rocket's trajectory should be as low as possible and the rocket should clear the building by at least 5m. What is the equation for the parabola formed by the trajectory. Graph the parabola.
Answer by Fermat(136)   (Show Source):
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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
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The green line is 35 m above the ground.
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