Answer:
Here, \[\frac{1}{2}m{{\upsilon
}^{2}}=1kJ=1000J\].
For maximum range, \[\theta
={{45}^{\text{o}}}\] and velocity of projectile at the highest point = \[\upsilon
\cos {{45}^{\text{o}}}\].
At the highest
point, \[KE=\frac{1}{2}m{{\left( \upsilon \cos {{45}^{\text{o}}}
\right)}^{2}}=\frac{1}{2}m\frac{{{\upsilon }^{2}}}{2}=\frac{1000}{2}=500J\]
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