# A projectile is shot at a velocity of 15 m/s and an angle of pi/12 . What is the projectile's maximum height?

Nov 5, 2017

0.77 m

#### Explanation:

sin(π/12) = (sqrt(3)-1) / (2sqrt(2))

H = (u^2 sin^2 θ)/(2g)

H = ((15ms^-1)^2 × sin^2(π/12))/(2 × 9.8 ms^-2)

H = ((15 ms^-1)^2 × ((sqrt(3)-1)/(2sqrt(2)))^2)/(2 × 9.8ms^-2)

H = (225 × (3 + 1 - 2sqrt(3))/(4 × 2))/(2 × 9.8)m

H = (225 × (4 - 2sqrt(3))/8)/(19.6)m = 0.77m (rounded to 2 significant figures)

Nov 5, 2017

The maximum height is 0.77 m.

#### Explanation:

Here,
${v}_{o} = 15 \frac{m}{s}$
${\Theta}_{o} = \frac{\pi}{12} \mathmr{and} {15}^{o}$
2D motion is made up of two independent 1D motion i.e vertical component and horizontal component.
Hence,
Vertical component be like,
${v}_{\text{0v}} = 15 \sin {\theta}_{0}$
${v}_{\text{0v}} = 15 \sin {15}^{o}$
$\therefore {v}_{\text{0v}} = 3.88 \frac{m}{s}$
We know at maximum height vertical velocity is equal to zero. So,
${V}_{\text{fv"=v_"0v}} - g \cdot t$
$O r , 0 = 3.88 \frac{m}{s} - 9.8 t$
Calculate the value of time (t) which is $0.39 s$
Which means at $0.39 s$ the projectile will achieve its maximum height.
So,
$y = {y}_{0} + \left({v}_{\text{0v}}\right) \cdot \left(t\right) - \left(1 / 2\right) g \cdot {t}^{2}$
$O r , y = 0 + \left(3.88 \frac{m}{s}\right) \left(0.39 s\right) - \left(1 / 2\right) \left(9.8 \frac{m}{s} ^ 2\right) {\left(0.39 s\right)}^{2}$
Calculate the value which is $0.77 m$ which is the maximum height.