Answer:
y maximum 3.54 m, value X 2.35 m
Explanation:
We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero
Vyf² = Vyo² - 2g (Y-Yo)
Where Yo is the initial height of the ramp 1.5 m
0 = Vyo² -2g (Y-Yo)
Y-Yo = Voy² / 2g
Y = Yo + Voy² / 2g
Let's calculate the velocity components using trigonometry
Voy = vo without T
Vox = Vo cost
Voy = 7.3 sin 60
Vox = 7.3 cos 60
Voy = 6.32 m / s
Vox = 3.65 m / s
Let's calculate the maximum height
Y = 1.5 +6.32²/2 9.8
Y = 3.54 m
This is the maximum height from the ground
b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point
Vfy = Voy - gt
0 = Voy - gt
t = Voy / g
t = 6.32 / 9.8
t = 0.645 s
Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time
X = Vox t
X = 3.65 0.645
X= 2.35 m
Answer:
13.2m
Explanation:
Step one:
given data
Energy= 5610J
Force F= 425N
Required
The distance traveled
Step two:
We know that work done is given as
WD= force* distance
so
5610=425*d
divide both sides by 425
d= 5610/425
d=13.2m
Answer:
a
Explanation:
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