Answer:
Maximum height = 2040 m
Explanation:
We can solve the problem using kinematics.
Consider the vertical motion of the object and use the equation:
[tex]\boxed{v^2 = u^2 + 2as}[/tex]
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 :
[tex]0^2 = (400 sin (30 \textdegree))^2 + 2(-9.81)(s)[/tex]
⇒ [tex]2(9.81)(s) = (400 sin (30 \textdegree))^2[/tex]
⇒ [tex]s = 2040 \space\ m[/tex] (3 s.f.)
