Answer:
a) Maximum height reached above ground = 2.8 m
b) When he reaches maximum height he is 2 m far from end of the ramp.
Explanation:
a) We have equation of motion v²=u²+2as
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0²=5.6² + 2 x -9.81 x s
s = 1.60 m
Height above ground = 1.2 + 1.6 = 2.8 m
b) We have equation of motion v= u+at
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0= 5.6 - 9.81 x t
t = 0.57s
Now considering horizontal motion of skateboarder.
We have equation of motion s =ut + 0.5 at²
u = 6.6cos58 = 3.50 m/s
a = 0 m/s²
t = 0.57
Substituting
s =3.5 x 0.57 + 0.5 x 0 x 0.57²
s = 2 m
When he reaches maximum height he is 2 m far from end of the ramp.
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In conduction, the heat transfer takes place within an object or between two objects in contact until the temperature becomes uniform. this kind of heat transfer continues until the temperature at two ends between which the heat transfer take place , becomes equal. Heat transfer takes place from point at high temperature to point at lower temperature.
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The conversion from gallons to liters is 1 = 3.785.
Keeping this in mind...
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B
The relatively less violent and calm nature of non-explosive volcanic eruptions can be modeled by squeezing a sponge. Option A would result in nothing while options C and D would result in a more violent explosion.
