1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.
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The total energy TE = mgh + 1/2 mU^2; where h = 20 m, g = 9.81 m/sec^2, and U = 10 mps. When the ball reaches max height H, all that TE will be potential energy PE = mgH = TE.
So there you are. TE = mgh + 1/2 mU^2 = mgH = TE from the conservation of energy. Solve for H.
1) H = (gh + 1/2 U^2)/g = h + U^2/2g = ? meters where everything on the RHS is given. You can do the math.
2) As the ball drops from H to h, it picks up KE as the potential energy mgH is converted when the potential energy is diminished to mgh, where h < H. So PE - pe = ke = mg(H - h) = 1/2 mv^2 so solve for v = sqrt(2g(H - h)) and, again, everything is given. You can do the math.
3) Same deal as 2) except now its V = sqrt(2gH) because all the PE = mgH = 1/2 mV^2 = KE when it is about to hit the ground. You can do the math.
Answer:
Yes
Explanation:
The displacement covered by a body in unit time is called velocity.
So here in order to find the average velocity we can say
so first we know that along the horizontal and along the inclined it moves with same time interval
so here we will have displacement in x direction as
now the average velocity in x direction will be given as
now similarly for y direction
first we will find its displacement
now the average velocity in y direction will be given as
now net velocity is given as
