A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an initial speed of $20 \mathrm{~ms}^{-1}$. If the ball strikes the ground after $3 \mathrm{~s}$. then the height of the building is (acceleration due to gravity $=10 \mathrm{~ms}^{-1}$ )

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Aicharya · Accepted Answer

Given situation is shown below

Here we are considering only vertical motion of ball. Given, initial vertical velocity of ball $ \begin{aligned} u_y & =u \sin \theta=20 \times \sin 30^{\circ} \\ & =20 \times \frac{1}{2}=10 \mathrm{~m} / \mathrm{s} \end{aligned} $ Vertical acceleration, $a_y=g=-10 \mathrm{~m} / \mathrm{s}^2$ Time to reach the ground $=3 \mathrm{~s}$. Now using, $h=u_y t+\frac{1}{2} a_y t^2$ We get, $h=10 \times 3+\frac{1}{2}(-10) \times 3^2$ $ \Rightarrow \quad h=30-45=-15 \mathrm{~m} \text {. } $ So height of building is $15 \mathrm{~m}$. (negative sign shows downward displacement of ball)

A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an…

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