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

Answer:

Maximum height = 2040 m

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.)