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A cannon ball is fired from the top of a $55 \mathrm{~m}$ high cliff with an initial speed of $50 \mathrm{~m} \mathrm{~s}^{-1}$. The speed of the cannon ball while hitting the ground in $\mathrm{m} \mathrm{s}^{-1}$ (acceleration due to. gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$60$
The given situation is shown below
According to work-energy theorern, Initial K. E of cannon ball + Gravitational P.E
$=$ Total K. E at ground
$\Rightarrow \quad \frac{1}{2} m v_1^2+m g h=\frac{1}{2} m v_2^2$
$\Rightarrow \quad v_1^2+2 g h=v_2^2 \Rightarrow v_2^2=v_1^2+2 g h$
Here, $v_1=$ initial speed $=50 \mathrm{~m} / \mathrm{s}$
$h=$ height of cliff $=55 \mathrm{~m}$
So, $\quad v_2^2=(50)^2+(2 \times 10 \times 55)$
$\begin{aligned} & \Rightarrow \quad v_2^2=2500+1100=3600 \\ & \text { or } \quad v_2=\sqrt{3600}=60 \mathrm{~m} / \mathrm{s}\end{aligned}$
According to work-energy theorern, Initial K. E of cannon ball + Gravitational P.E
$=$ Total K. E at ground
$\Rightarrow \quad \frac{1}{2} m v_1^2+m g h=\frac{1}{2} m v_2^2$
$\Rightarrow \quad v_1^2+2 g h=v_2^2 \Rightarrow v_2^2=v_1^2+2 g h$
Here, $v_1=$ initial speed $=50 \mathrm{~m} / \mathrm{s}$
$h=$ height of cliff $=55 \mathrm{~m}$
So, $\quad v_2^2=(50)^2+(2 \times 10 \times 55)$
$\begin{aligned} & \Rightarrow \quad v_2^2=2500+1100=3600 \\ & \text { or } \quad v_2=\sqrt{3600}=60 \mathrm{~m} / \mathrm{s}\end{aligned}$
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