Answer:
1.) 17.3 m/s
2.) 1.8 s
Step-by-step 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.