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??

Answer 1

#t = 0.422 s# (3 sf)

We can use an equation of constant acceleration to solve for the time by first figuring out the vertical component of the velocity in order to answer this question.

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

Apply the following equation for constant acceleration: NB The vertical component of velocity equals zero at maximum height, and I will interpret upward 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 + at#

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

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Answer 2

Xavier Adame

The time to reach the maximum height can be calculated using the formula: (t = \frac{V_y}{g})

Where: (V_y) is the vertical component of the initial velocity ((V_y = V \cdot \sin(\theta))), (g) is the acceleration due to gravity.

(t = \frac{16 , \text{m/s} \cdot \sin(\pi/12)}{9.8 , \text{m/s}^2})

(t \approx 0.92 , \text{s})

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.