Answer:
t = 123.59s
Explanation:
For the launch pad section:
Vf = Vo + a*t where Vo=0.
Vf = 35*25 = 875m/s
The distance traveled during the launch:

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

where d= 10937.5m; Vo=875m/s.
Solving for t:
t1 = -11.093s t2 = 98.59s
So, the total time of flight will be:

Answer:
90 m/s
8100 m
Explanation:
Given:
v₀ = 0 m/s
a = 0.5 m/s²
t = 3 min = 180 s
Find: v and Δx
v = at + v₀
v = (0.5 m/s²) (180 s) + 0 m/s
v = 90 m/s
Δx = v₀ t + ½ at²
Δx = (0 m/s) (180 s) + ½ (0.5 m/s²) (180 s)²
Δx = 8100 m
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