Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Screenshot_2019-03-02-15-01-13-335_com.android.chrome.png A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v , the total area around the fountain that gets wet is

Answers (1)

Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

R=\frac{u^{2}\sin 2\Theta }{g}

- wherein

Special case of horizontal range

For max horizontal range.

\Theta = 45^{0}

R_{max}=\frac{u^{2}\sin 2 (45) }{g}=\frac{u^{2}\times 1}{g}=\frac{u^{2}}{g}

 

 R_{max}=\frac{v^2_0sin 90^{o} }{g}=\frac{v^2_0}{g}

Area=\Pi (R_{max})^2 \: =\frac{\Pi v^4_0}{g^2}

Posted by

Safeer PP

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 Registration LIVE: NTA JEE form at jeemain.nta.nic.in; what's new this year?
anam-khan

JEE Main 2026: Wider gap in NIT cut-offs for home and other-state quotas; outsiders need better rank
anam-khan

JEE Mains 2026: Use of calculator 'not allowed’ says NTA; cites ‘typographical error' in information bulletin

Latest Question