Arocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/s2. after 25.0 s of powered flight, the engines shut off and the rocket follows a parabolic path back to earth. find the time of flight from launch to impact. hint: simple projectile motion after engines are shut down.

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:

Explanation:

The answer to your question would be number 2 amplitude and frequency are the same so the higher the frequency the higher the amplitude. As you can see in the picture wave E has the  highest frequency. which means since A has the lowest  frequency and E has the highest frequency answer 2 would be correct the amplitude increases from A to E.