Answer:

Explanation:
Let the distance from spotlight to wall be 15m, and distance from the man to the building be
.
#Therefore the height of the shadow as a function of the above is 
Hence, height of the shadow is expressed as s=(15-x)m
#See attached photo for illustration
Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.
Now here we will do the components of the weight of the person
given that weight of the person = 500 N
now its components are


now here as we can say that one of the component is balanced here by the normal force perpendicular to plane
while the other component of the weight is balanced by the force applied on the rope
So here the force applied on the rope will be given as


so it apply 300 N force along the inclined plane
Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:
Vv=80*sin(10)=80*0.1736=13.888 m/s.
So the vertical component of the velocity of the train is Vv=13.888 m/s.