Answer:
44.64 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
<u>Time taken to reach 1180 m is 11.29 seconds</u>
<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>
The distance the rocket will keep going up after the engines shut off is 836.05 m
Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m
The rocket will fall from this height
<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>
Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds
<span>1078 kgm / s would be the answer I hope this helps!!!</span>
I think that the answer is friction
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Torque acting dowward = 6 x 0.5 = 3 Nm
Torque acting to the right = 5 x 1 = 5 Nm
5 - 3 = 2 Nm
inertia = 1/2 mr^2
0.5 x 10 x 1^2 = 5 kg-m^2
2/5 = alpha = 0.4 rad /s^2
Hope this helps
