d=vi*t+(1/2)gt²
d=11 m
g=9.8 m/s²
vi=0 m/s
11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s
Answer: the time the sone is in flight is 1.5 s
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
Answer:
1. 218.55 N
2. 
3. 
Explanation:
Part 1;
Net force
where m is mass, g is gravitational force and
is the angle of inclination

Frictional force,
is given by
where
is the coefficient of static friction


Since
, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N
Part 2.
Using the relationship that
Frictional force 



The maximum angle of inclination 

Part 3:
Net force on the object is given by
where
is the coefficient of kinetic friction

= 9.8 ( sin 38 - (0.51) cos 38 )
= 