Question

A first year student projected a farm business brochure to a farmer at 30 degrees to horizontal. calculate the maximum height attained by the projectile if it was launched at 400m/s​

Answer 1

Maximum height = 2040 m

Step-by-step explanation:

We can solve the problem using kinematics.

Consider the vertical motion of the object and use the equation:

`\boxed{v^2 = u^2 + 2as}`

where:

• v = final velocity (0 m/s, because when the object is at max. height, it has no vertical velocity)

• u = initial velocity (400sin30° m/s ⇒ vertical component of 400 m/s at 30° to horizontal)

• a = acceleration (-9.81 m/s²; considering upward acceleration to be negative)

• s = displacement (? m; this represents the max. height of the object),

Substitute the values into the equation and solve for s :

`0^2 = (400 sin (30 \textdegree))^2 + 2(-9.81)(s)`

⇒
`2(9.81)(s) = (400 sin (30 \textdegree))^2`

⇒
`s = 2040 \space\ m` (3 s.f.)