Question

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?

Answer 1

The time is `=9.6s`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=47*sin(1/2pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=47sin(1/2pi)-g*t`

`t=47*1/g*sin(1/2pi)`

`=4.8s`

The time for the projectile to land is

`=2*4.8=9.6s`

Answer 2

`9.8s`

Explanation:

If you carefully look into the question:

`pi/2=180/2=90^circ`

`:.` 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`

`rarrv=u+at`

`rarr0=47-9.8t`

`rarr9.8t=47`

`rarrt=47/9.8`

`color(green)(rArr~~4.8`

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

So, `"Total time"=4.8*2=9.8s`

Hope this helps.. :)