A ball of mass 2 ,kg is thrown from a tall bu | vedclass

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Question

(A) Given: mass $m = 2 \,kg$, initial velocity $\vec{u} = (20 \hat{i} + 24 \hat{j}) \,m/s$.
The vertical component of velocity at time $t$ is given b

Vedclass · Accepted Answer

$-2.56$

Vedclass · Answer

(A) Given: mass $m = 2 \,kg$, initial velocity $\vec{u} = (20 \hat{i} + 24 \hat{j}) \,m/s$.
The vertical component of velocity at time $t$ is given by $v_y = u_y - gt$.
At $t = 8 \,s$, $v_y = 24 - (10 	imes 8) = 24 - 80 = -56 \,m/s$.
The horizontal component remains constant: $v_x = 20 \,m/s$.
So, final velocity $\vec{v} = (20 \hat{i} - 56 \hat{j}) \,m/s$.
By the Work-Energy Theorem, the change in potential energy $\Delta PE$ is equal to the negative of the change in kinetic energy $\Delta KE$.
$\Delta PE = -\Delta KE = -(KE_f - KE_i) = KE_i - KE_f$.
$\Delta PE = \frac{1}{2} m (u^2 - v^2) = \frac{1}{2} 	imes 2 	imes [(20^2 + 24^2) - (20^2 + (-56)^2)]$.
$\Delta PE = (400 + 576) - (400 + 3136) = 976 - 3536 = -2560 \,J$.
Converting to kilojoules, $\Delta PE = -2.56 \,kJ$.

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