you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point
1 answer:
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Answer:
v = 17.30 m / s
Explanation:
For this exercise we will use Newton's second law
at the bottom of the loop and stopped
∑ F = 0
N-W = 0
N = W
W = 770 N
the mass of the body is
W = mg
m = W / g
m = 770 / 9.8
m = 78.6 kg
on top of the loop and moving
∑ F = m a
N + W = m a
note that the three vectors go in the same vertical direction down
the centripetal acceleration is
a = v² / r
we substitute
N + W = m v² / r
v =
let's calculate
v =
v = 17.30 m / s