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:
[tex]d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m[/tex]
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.
[tex]\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}[/tex]
[tex]-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}[/tex] where d= 10937.5m; Vo=875m/s.
Solving for t:
t1 = -11.093s t2 = 98.59s
So, the total time of flight will be:
[tex]t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s[/tex]
