A particle is projected up from a point at an angle \(\theta\) with the horizontal direction. At any time \(t\), if \(p\) is the linear momentum, \(y\) is the vertical displacement, \(x\) is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy \(KE\) of the particle is
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For projectile motion, KE is minimum at the highest point and decreases linearly with height: \(KE = KE_0 - mgy\).
Step 1: General kinetic energy relation.
Kinetic energy is:
\[
KE = \frac{p^2}{2m}
\]
So, \(KE\) is proportional to \(p^2\). Step 2: KE vs \(p^2\).
Since \(KE \propto p^2\), graph (D) showing linear relation is correct. Step 3: KE vs time.
In projectile motion, speed reduces up to top point and then increases again, so KE vs time is a U-shaped curve.
So graph (B) is correct. Step 4: KE vs vertical displacement \(y\).
As particle rises, KE decreases linearly with increase in height because potential energy increases,
\[
KE = KE_0 - mgy
\]
So KE vs \(y\) should be a straight line with negative slope, not a V-shape.
Graph (A) shows a V-type behavior, which is not physically correct. Step 5: KE vs horizontal displacement \(x\).
In projectile motion, speed depends on time, and \(x\) increases linearly with time, so KE vs \(x\) is also U-shaped.
So graph (C) is acceptable. Final Answer:
\[
\boxed{\text{graph (A)}}
\]
