<span>ANSWER: the height is same either way and hang time is same either way.
EXPLANATION :
suppose if a basketball player jumps vertically will have a hang time around half a second. The remaining half a second he spends in the top of the jump.</span>
Answer: 60m/s
Explanation:
From the diagram:
Θ = 30°
Vertical resolution (y-axis) :
Voy = VoSinΘ
g in the upward direction = negative (-) = - g
Vfinal = 0
Distance (H) traveled along y =
Time taken to reach maximum height :
From v = u + at
0 = usinΘ - gt
gt = usinΘ
t = usinΘ / g
Horizontal resolution:
S = ut + 1/2at^2
Substituting t = usinΘ / g ; Voy = usinΘ
S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2
S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)
S = (u^2sin^2Θ) / 2g
Now if S = maximum height = 45m
Then,
45 = [Vo^2sin^2(30°)] / 2(10)
45 =[ Vo^2 * (0.5)^2] / 20
45 =( Vo^2 * 0.25) / 20
20 * 45 = Vo^2 * 0.25
900 / 0.25 = Vo^2
3600 = Vo^2
Vo = sqrt(3600)
Vo = 60m/s
Answer:
Explanation:
Given
Mass of first object 
Mass of second object 
Distance between them 
object is placed between them
So force exerted by
on 



Force exerted by 



So net force on
is



i.e. net force is towards 
(b)For net force to be zero on
, suppose
So force exerted by
and
must be equal







