If a projectile is shot at an angle of pi/12 and at a velocity of 16 m/s, when will it reach its maximum height??

Jan 21, 2016

$t = 0.422 s$ (3 sf)

Explanation:

To solve this question we will first work out the vertical component of the velocity, then we can apply an equation of constant acceleration to the vertical motion to solve for the time.

Calculate the vertical component of the velocity:
u_V = u sin (θ) = 16 × sin (π//12) = 4.141 m.s^(-1)

Apply an equation of constant acceleration:
NB At maximum height the vertical component of velocity is equal to zero. And I will take upwards as positive .

List all known values:
s_V = ?
${u}_{V} = 4.141 m . {s}^{- 1}$
${v}_{V} = 0$
$a = - 9.81 m . {s}^{- 2}$ (Negative because it acts downwards)
t = ?

Use ${v}_{V} = {u}_{V} + a t$

Rearrange for t :
⇒ t = (v_V-u_V)/a = (0-4.141)/-9.81 = 0.422 s (3 sf)