A ball is dropped from a distance 5 m above the ground, and it hits the ground with a certain speed. If the same ball is dropped from a distance 10 m above the ground, its final speed will be A ball is dropped from a distance 5 m above the ground, and it hits the ground with a certain speed. If the same ball is dropped from a distance 10 m above the ground, its final speed will be twice as high. the same. 1.4 times as high.

1.4 times as high

Step-by-step explanation:

Now here we have to compare the both final velocities by using the formula independent of time. As we know,

2aS = V(f)2 - V(i)2

Both balls are dropped from certain height, so V(i) = 0 m/s for both balls

So, 2aS = V(f)2 for both balls.

Now if we compare both velocities

Equation for first ball

2aS(1) = V(1)2

Equation for second ball

2aS(2) = V(2)2

Comparing both equations

2aS(1)/2aS(2) = V(1)2/V(2)2

as we now S(1) = 5m and S(2) = 10m, also acceleration a = g, because acceleration due to gravity is acting on the balls. So,

2(9.8) x 5/2(9.8) x 10 = V(1)2/V(2)2

after simplifying,

1/2 = V(1)2/V(2)2

V(2)^2 = 2V(1)^2

By taking square root on both sides,

V(2) = 1.4V(1)

step-by-step explanation:

for the free

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh