Question

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)?

Answer 1

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Answer 2

"According to the given data, velocity vector is given from which we get the velocity

$v=\sqrt { 8\times 18+24\times 24 }$

$\Rightarrow v=30\frac { m }{ s }$

Thereby the distance travelled in 6 seconds is given by the equation of motion,

$s=ut+\frac { 1 }{ 2 } { at }^{ 2 }$

$\Rightarrow s=30\times 6+\frac { 1 }{ 2 } \times 10\times 6\times 6$

$\Rightarrow s=360\quad m$

Therefore, the initial potential energy be

$U_i = mgh = 1 \times 10 \times h$

Final potential energy be

$U_f = 1 \times 10 \times ( h - 360)$

$\Rightarrow thereby\quad \delta U = U_f - U_i$

$\Rightarrow \delta U = 1 \times 10 \times ( h - 360) - 10 h$

$\Rightarrow \delta U = -3600\quad Joules$"