the answer to the question is (D)
Answer:
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
Explanation:
The average speed is defined by the variation of the position between the time spent
v = Δx / Δt
since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is
Δx = 20 - 15 +20
Δx = 25 m
therefore we calculate
v = 25/75
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
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Answer:
48.26 m
Explanation:
time to goes up (till stop for a while in the air - maximum height)
vt = vo + a t
0 = 15 + g . t
0 = 15 + (-9.8) . t
9.8t = 15
t = 1.531 s
so the time left to goes down is
4.0 - 1.531 = 2.469 s
height from the top of building can find it by using
vo =√(2gh)
15 = √(2)(9.8).h
15² = 19.6h
h = 225/19.6 = 11.48 m
so the distance of maximum height to the ground is
t = √(2H/g)
2.469 = √(2H/9.8)
2.469² = 2H/9.8
6.096 = 2H/9.8
2H = 6.096 x 9.8 = 59.74 m
so the vertical distance of the building (or the building height's is)
H - h = 59.74 - 11.48 = 48.26 m
