A projectile fired at 30 ° to the ground is observed to be at same height at time 3s and 5s after projection, during its flight. The speed of projection of the projectile is m s ^ - 1 . (Given g = 10m s ^ - 2 )

Question

Quantrex Academy · Accepted Answer

Correct answer: 80

A projectile fired at 30 ° to the ground is observed to be at same height at time 3s and 5s after projection, during

The time taken to reach maximum height is

$T ' = \dfrac{usin\theta}{g}$ .

The total time of flight is $T = ( 3 + 5 ) s = 8s$ .

Hence,

$T ' = \left(\dfrac{T}{2}\right) = \left(\dfrac{3 + 5}{2}\right) = \left(\dfrac{usin\theta}{g}\right)$

$⇒ 4 = \dfrac{( u ) \times \dfrac{1}{2}}{10}$

$⇒ u = 80m{s}^{- 1}$

A projectile fired at 30 ° to the ground is observed to be at same...

Correct answer

Step-by-step solution

More from Motion in Two Dimensions