While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.25 m/s. The stone subsequently falls to the ground, which is 13.3 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.8 m/s2.

1.) 17.3 m/s

2.) 1.8 s

Explanation:

Given that While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.25 m/s.

The maximum height will be calculated by using third equation of motion

At maximum height, V = 0

V^2 = U^2 - 2gH

0 = 6.25^2 - 2 × 9.8 × H

39.06 = 19.6H

H = 39.06 / 19.6

H = 1.993m

The stone subsequently falls to the ground, which is 13.3 m below the point where the stone leaves your hand. At what speed does the stone impact the ground?

Using the same formula again.

Where the height = 13.3 + 1.993

Height = 15.29m

Substitutes the height into the third equation of motion

V^2 = U^2 + 2gH

U = 0

V^2 = 0 + 2 × 9.8 × 15.29

V^2 = 299.74

V = sqrt ( 299.74 )

V = 17.3 m/s

How much time is the stone in the air? Ignore air resistance and take g = 9.8 m/s2.

Using the second equation of motion formula

H = Ut + 1/2gt^2

U = 0

15.29 = 1/2 × 9.8 × t^2

15.29 = 4.9t^2

t^2 = 15.29 / 4.9

t^2 = 3.120

t = sqrt ( 3.120 )

t = 1.766 s

Therefore, the stone is on air for 1.8s approximately.

good for her? ? where is the question?

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures