(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²
Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
The answer is D. i got all the other ones wrong ._.
Answer:
v = -1.8t+36
20 seconds
360 m
40 seconds
36 m/s
The object speed will increase when it is coming down from its highest height.
Explanation:

Differentiating with respect to time we get

a) Velocity of the object after t seconds is v = -1.8t+36
At the highest point v will be 0

b) The object will reach the highest point after 20 seconds

c) Highest point the object will reach is 360 m


d) Time taken to strike the ground would be 20+20 = 40 seconds
![[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s](https://tex.z-dn.net/?f=%5Btex%5Dv%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B0.9%5Ctimes%202%5Ctimes%2020%5C%5C%5CRightarrow%20v%3D36%5C%20m%2Fs)
Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of 
e) The velocity with which the object strikes the ground will be 36 m/s
f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.
Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N