Courses for Kids
Free study material
Offline Centres
Store Icon

The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall?

Last updated date: 09th Apr 2024
Total views: 33.6k
Views today: 0.33k
MVSAT 2024
33.6k+ views
Hint:- Proceed the solution of this question, with clear understanding of what is given in question, as maximum height and initial velocity is given so we can use the formula of maximum height of projectile motion from there we can get the value of $\sin \theta {\text{ & cos}}\theta $ which is required in the formula of horizontal range.

Complete step-by-step solution -

In the question it is given that,
Initial velocity u= 40 m/s
Maximum height H =25 m
Consider $g = 10{\text{ }}\dfrac{{\text{m}}}{{{{\sec }^2}}}$
we know that he maximum height in projectile motion equal to $\dfrac{{{{\text{u}}^2}{\text{si}}{{\text{n}}^2}\theta }}{{2g}}$
so equalising it with given maximum height in question i.e. 25 m
$ \Rightarrow \dfrac{{{{\text{u}}^2}{\text{si}}{{\text{n}}^2}\theta }}{{2g}} = 25$
Hence find the value of ${\text{sin}}\theta $ from here
$ \Rightarrow {\text{si}}{{\text{n}}^2}\theta = \dfrac{{25 \times 2g}}{{{{\text{u}}^2}}}$
$ \Rightarrow {\text{si}}{{\text{n}}^2}\theta = \dfrac{{25 \times 20}}{{1600}} = \dfrac{{500}}{{1600}} = \dfrac{5}{{16}}$
$ \Rightarrow {\text{sin}}\theta = \sqrt {\dfrac{5}{{16}}} $ ………….(1)
$ \Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \dfrac{5}{{16}}} = \sqrt {\dfrac{{11}}{{16}}} $ …………(2)

We know that the maximum horizontal distance will be equal to the horizontal range in projectile motion.
$ \Rightarrow {\text{Range (R) = }}\dfrac{{{{\text{u}}^2}\sin 2\theta }}{g}$
Using $\sin 2\theta = {\text{ }}2\sin \theta \cos \theta $
$ \Rightarrow {\text{R = }}\dfrac{{{{\text{u}}^2}\left( {2\sin \theta \cos \theta } \right)}}{g}$
Hence on putting the values of $\sin \theta {\text{& cos}}\theta $ from expression (1) and (2)
$ \Rightarrow {\text{R = }}\dfrac{{1600\left( {2\sqrt {\dfrac{5}{{16}}} \times \sqrt {\dfrac{{11}}{{16}}} } \right)}}{{10}} = 160 \times 2\dfrac{{\sqrt {55} }}{{16}} = 20 \times \sqrt {55} = 148.32$

Therefore, Range (R) = 148.32 m

Note- In the question of projectile motion, it is advisable to remember some general formulas of projectile motion as formula of maximum height, Range and Time of flight, then this way we can save our time otherwise it will be lengthier to derive every time and chances of mistakes also increases.