Question

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

Answer 1

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)

Answer 2

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.