At t=0 a 1.0kg ball is thrown from a tall tower with v=(18 m/s)i+(24 m/s)j. what is δu of the ball-earth system between t=0 and t=6.0s (still free fall)δ

ΔU = 3492.72 J

Explanation:

Given

m = 1.0 kg

v = 18 i m/s + 24 j m/s

To find the initial distance to determine the initial potential energy can use newton's equations

Now the potential energy can be find using

and final energy as t=6

Initial and final energy can be related as:

ΔU = m*g*h - m*g*(h-356.4m)

So get the factor to solve the total energy

ΔU = m*g * (h - h + 356.4)

ΔU = 1.0 kg * 9.8  m/s^2 * 356.4 m

ΔU = 3492.72 J

(75°f − 32) × 5/9 = 23.889°c

c. newton's third law of motion