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:
Mass doesn't change, no matter where you take it.
Your first impression of ' 0 ' is totally correct.
Well if the boat initially at rest accelerates at uniformly at 4.0 m/s (squared) then it would be best to muitlply it so 4.0 squared equals 2 by multiplying that by 7.0 your answer would be 14 s
Answer and Explanation:
The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence
For example let the initial height is 243 fit
After one bounce it will reach 243/3 =81 feet
After second bounce 81/3=27 feet
After third bounce 27/3 =9 feet
After fourth bounce 9/3 =3 feet
So a sequence is formed that is 243,81,27,9,3..........
Here
Sum of infinite GP =
From this formula we can find the total distance traveled by the ball
<span>C. Mao Zedong
Hope this helps!~</span>