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?

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Answer 1 · 2018-06-27T07:25:15

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

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

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

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.

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

A projectile is shot from the ground at a velocity of 8 m/s8 m/s and at an angle of (2pi)/3(2pi)/3. How long will it take for the projectile to land?