Question

At t=0 a 1.0 kg ball is thrown from the top of a ball tower with velocity y = (8î+9j) m/s. The change in the potential energy of the ball-earth system between t=0 to t= 5s: (g= 10 m/s)​

Answer 1

Answer:

Given,

m=1 kg

∣v∣=30 m/s

t=6sec

Since time does not directly enter into the energy formulations,

we return to Chapter 4 (or Table 2−1 in chapter 2) to find the change of height during this t=6.0s flight.

Δy=v

0y

t−

2

1

gt

2

This leads to Δy=−32m.

Therefore ΔU=mgΔy−318J≈−3.2×10

−2

J.

Answer 2

Answer:

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Step-by-step explanation:

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