A NCC parade is going at a uniform speed of 9 km / h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given .g=9.8 m / s 2)

Question

A NCC parade is going at a uniform speed of 9 km / h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given .g=9.8 m / s 2) (A) 5 m (B) 10 m (C) 19.8 m (D) 24.5 m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8c1f1c1c1798af031b9193a7a599e55e-.png" /> Time taken by mango =√(2 n / g ) =√(2 × 19.6/9.8)=2 second Distance = vt =9 × (5/18) × 2=5 m

Q. A NCC parade is going at a uniform speed of $9 km / h$ under a mango tree on which a monkey is sitting at a height of $19.6 m$ . At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is : (Given $g = 9.8 m / s^{2})$

$5 m$

$10 m$

$19.8 m$

$24.5 m$

Time taken by mango $= \frac{2 n}{g}$
$= \frac{2 \times 19.6}{9.8} = 2$ second
Distance $= v t$
$= 9 \times \frac{5}{18} \times 2 = 5 m$

Q. A NCC parade is going at a uniform speed of 9km/h under a mango tree on which a monkey is sitting at a height of 19.6m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is : (Given g=9.8m/s2)

5m

10m

19.8m

24.5m