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

Question

A projectile fired at 30 &#176; to the ground is observed to be at same height at time 3 &#160; s and 5 &#160; s after projection, during its flight. The speed of projection of the projectile is m &#160; s - 1 . (Given g = 10 &#160; m &#160; s - 2 )

Aicharya · Accepted Answer

The time taken to reach maximum height is

$T' = \frac{u \sin θ}{g}$ .

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

Hence,

$T' = (\frac{T}{2}) = (\frac{3 + 5}{2}) = (\frac{u \sin θ}{g})$

$\Rightarrow 4 = \frac{(u) \times \frac{1}{2}}{10}$

$\Rightarrow u = 80 m s^{- 1}$

A projectile fired at 30 ° to the ground is observed to be at same height at time 3   s and 5…

Solution

A projectile fired at 30 &#176; to the ground is observed to be at same height at time 3 &#160; s and 5…

Solution

A projectile fired at 30 ° to the ground is observed to be at same height at time 3 s and 5…