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

2 Answers
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=15m/s
Theta_o=pi/12 or 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_"0v"=15sintheta_0
v_"0v"=15sin15^o
:.v_"0v"=3.88m/s
We know at maximum height vertical velocity is equal to zero. So,
V_"fv"=v_"0v"-g*t
Or, 0=3.88m/s-9.8t
Calculate the value of time (t) which is 0.39s
Which means at 0.39s the projectile will achieve its maximum height.
So,
y=y_0+(v_"0v")*(t)-(1//2)g*t^2
Or, y=0+(3.88m/s)(0.39s)-(1//2)(9.8m/s^2)(0.39s)^2
Calculate the value which is 0.77m which is the maximum height.