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A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of 15 $\mathrm{ms}^{-1}$. If the ball hits the ground after $3 \mathrm{~s}$, then the height of building is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$22.5 \mathrm{~m}$
$\frac{15}{2} \mathrm{~m} / \mathrm{s}$
Time taken by ball to reach the same height of building,
$\mathrm{t}=\frac{2 \mu \sin \theta}{\mathrm{g}}=\frac{2 \times 15 \times \sin 30^{\circ}}{10}=1.5 \mathrm{sec}$
Now, remaining $1.5 \mathrm{sec}$ will be taken by ball to reach the ground a Considering only vertical motion of ball
$\begin{aligned} & \mathrm{h}=\frac{15}{2} \times 1.5+\frac{1}{2} \times 10 \times 1.5^2 \\ & =22.5 \mathrm{~m}\end{aligned}$
Time taken by ball to reach the same height of building,
$\mathrm{t}=\frac{2 \mu \sin \theta}{\mathrm{g}}=\frac{2 \times 15 \times \sin 30^{\circ}}{10}=1.5 \mathrm{sec}$
Now, remaining $1.5 \mathrm{sec}$ will be taken by ball to reach the ground a Considering only vertical motion of ball
$\begin{aligned} & \mathrm{h}=\frac{15}{2} \times 1.5+\frac{1}{2} \times 10 \times 1.5^2 \\ & =22.5 \mathrm{~m}\end{aligned}$
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