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A ball is tossed up in the air. at its peak, it stops before beginning to fall. the ball at its peak has

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The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.

so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0

The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.
User Brnrd
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Answer:potential energy only?

Explanation: the ball is at the top which stores energy like potential energy

User Googol
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