Question

A projectile is shot from the ground at a velocity of `8 m/s` and at an angle of `(2pi)/3`. How long will it take for the projectile to land?

Answer 1

`|vecv| = 8m*s^(-1)`

`vecv_x = 8cos((2pi)/3) m*s^(-1) hati=-4 hati`

`vecv_y = 8sin((2pi)/3) m*s^(-1) hatj=4sqrt3hatj`

In y axis,

`v_i = 4sqrt3`

At maximum height,

`v_f=0`

`a=-10 m*s^(-2)`

Applying first equation of motion,

`0=4sqrt3 - 10t rArr t=0.4*sqrt3 rArr t=0.692s`

So, time taken in reaching maximum height is 0.692 s.Time taken for projectile to fall from maximum height to ground is also 0.692s.

`"Total time taken" = 2*0.692 = 1.384s`