The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
The magnitude of the resultant is
√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .
The direction of the resultant is
tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .
I'm pretty sure that it's 815.
Answer: c. A person who gains unauthorized access to digital data
Explanation:
