A projectile fired at 30^{circ} to the ground | vedclass

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(B) For a projectile,the time of flight $T$ is the total time taken to return to the ground. If a projectile is at the same height $h$ at two differen

Vedclass · Accepted Answer

$80$

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(B) For a projectile,the time of flight $T$ is the total time taken to return to the ground. If a projectile is at the same height $h$ at two different times $t_1$ and $t_2$,then the total time of flight $T$ is given by $T = t_1 + t_2$.Given $t_1 = 3 \, s$ and $t_2 = 5 \, s$,the total time of flight is $T = 3 + 5 = 8 \, s$.The formula for the time of flight is $T = \frac{2 u \sin 	heta}{g}$,where $u$ is the initial speed,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.Substituting the given values: $8 = \frac{2 u \sin 30^{\circ}}{10}$.Since $\sin 30^{\circ} = 0.5$,we have $8 = \frac{2 u (0.5)}{10}$.$8 = \frac{u}{10}$.Therefore,$u = 80 \, m \, s^{-1}$.

$A$ projectile fired at $30^{\circ}$ to the ground is observed to be at the same height at time $t_1 = 3 \, s$ and $t_2 = 5 \, s$ after projection,during its flight. The speed of projection of the projectile is $......... \, m \, s^{-1}$ (Given $g = 10 \, m \, s^{-2}$).

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