TS EAMCET · Physics · Work Power Energy
A ball of mass \(m=1 \mathrm{~kg}\) is thrown from the top of a building with initial velocity \(\mathbf{v}=(20 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(24 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\) at time \(t=0\). The change in the potential energy of the ball between \(t=0\) and \(t=6 \mathrm{~s}\), if the ball does not hit the ground, then (assume, \(g=10 \mathrm{~m} \mathrm{~s}^2\) )
- A \(-320 \mathrm{~J}\)
- B \(-360 \mathrm{~J}\)
- C \(-380 \mathrm{~J}\)
- D \(320 \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(-360 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(\mathbf{u}=20 \hat{\mathbf{i}}+24 \hat{\mathbf{j}}\) To find change in PE, we have to consider only vertical motion of the ball. So, we have \(u_y=24 \mathrm{~m} / \mathrm{s}, m=1 \mathrm{~kg}\) \(a_y=-10 \mathrm{~m} / \mathrm{s}^2, t=6 \mathrm{~s}\) We take origin at point of…
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