# A projectile is shot from the ground at a velocity of 47 m/s and at an angle of (pi)/2. How long will it take for the projectile to land?

Apr 11, 2017

The time is $= 9.6 s$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 47 \cdot \sin \left(\frac{1}{2} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = 47 \sin \left(\frac{1}{2} \pi\right) - g \cdot t$

$t = 47 \cdot \frac{1}{g} \cdot \sin \left(\frac{1}{2} \pi\right)$

$= 4.8 s$

The time for the projectile to land is

$= 2 \cdot 4.8 = 9.6 s$

Apr 11, 2017

$9.8 s$

#### Explanation:

If you carefully look into the question:

$\frac{\pi}{2} = \frac{180}{2} = {90}^{\circ}$

$\therefore$ The projectile moves up and comes down vertically

So, To solve, we use

color(blue)(v=u+at

Where,

color(orange)(v="Final velocity"=0

color(orange)(u="Initial velocity"=47 m/s

color(orange)(a="Acceleration"=-9.8 (As, gravity pulls down)

color(orange)(t="Time" (We need to find it)

Let's solve for $t$

$\rightarrow v = u + a t$

$\rightarrow 0 = 47 - 9.8 t$

$\rightarrow 9.8 t = 47$

$\rightarrow t = \frac{47}{9.8}$

color(green)(rArr~~4.8

Now we have found the time for the projectile to reach its maximum height

So, $\text{Total time} = 4.8 \cdot 2 = 9.8 s$

Hope this helps.. :)