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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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(acceleration due to gravity $=10 \mathrm{~ms}^{-1}$ )
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Verified Answer
The correct answer is:
$15 \mathrm{~m}$
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).
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).
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