# A projectile is shot from the ground at a velocity of 15 m/s at an angle of pi/6. How long will it take for the projectile to land?

Dec 17, 2017

The time is $= 1.53 s$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

The initial velocity is ${u}_{0} = 15 \sin \left(\frac{1}{6} \pi\right) m {s}^{-} 1$

The time to reach the greatest height is $= t s$

Applying the equation of motion

$v = u + a t = u - g t$

The time is $t = \frac{v - u}{- g} = \frac{0 - 15 \sin \left(\frac{1}{6} \pi\right)}{- 9.8} = 0.765 s$

The time taken to land is twice the time to reach the greatest height

$T = 2 t = 2 \cdot 0.765 = 1.53 s$

Dec 17, 2017

$1.531 s$

#### Explanation:

The initial velocity,${v}_{0} = 15 m {s}^{- 1}$
The throwing angle of the projectile,θ=$\frac{\pi}{6} = 30 \circ$
$S u p p o s e ,$
The wandering period of the projectile, color(red)(T)=?
$T = \textcolor{b l u e}{\frac{2 {v}_{0} \sin \theta}{g}} = \frac{2 \times 15 m {s}^{- 1} \times \sin 30 \circ}{9.8 m {s}^{- 2}} = 1.531 s$color(red)((Ans.)