Search any question & find its solution
Question:
Answered & Verified by Expert
A player can throw a ball to a maximum horizontal distance of $80 \mathrm{~m}$. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is
Options:
Solution:
2930 Upvotes
Verified Answer
The correct answer is:
$40 \mathrm{~m}$
The maximum horizontal distances is given by
$$
\begin{aligned}
& \mathrm{H}_{\max }=\frac{\mathrm{u}^2}{\mathrm{~g}} \sin 90^{\circ} \Rightarrow 80=\frac{\mathrm{u}^2}{\mathrm{~g}} \\
& \mathrm{u}^2=80 \mathrm{~g}
\end{aligned}
$$
The ball will achieve the maximum height when it is thrown vertically upward.
$$
\begin{aligned}
& v^2=u^2-2 g h \\
& 0=80 g-2 g h \Rightarrow h=40 m
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{H}_{\max }=\frac{\mathrm{u}^2}{\mathrm{~g}} \sin 90^{\circ} \Rightarrow 80=\frac{\mathrm{u}^2}{\mathrm{~g}} \\
& \mathrm{u}^2=80 \mathrm{~g}
\end{aligned}
$$
The ball will achieve the maximum height when it is thrown vertically upward.
$$
\begin{aligned}
& v^2=u^2-2 g h \\
& 0=80 g-2 g h \Rightarrow h=40 m
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.