Question

example 2 - free fall

a stone is thrown upward with an initial velocity of 30 m/s.
a) how high above the point of release will the stone rise?
b) what will be the velocity of the stone when it returns to your hand?
c) for how long will the stone remain in the air?

Explanation:

(a):

Step1: Known quantities and equation

u=30 m/s, v=0 m/s, a=-9.8 m/s², use \(v^2 = u^2 + 2as\)

Step2: Calculate height

\(s=\frac{v^2 - u^2}{2a}=\frac{0 - 30^2}{2\times(-9.8)}=\frac{-900}{-19.6}\approx45.92\ \text{m}\)

(b):

Step1: Free fall symmetry

Return speed = initial speed, direction opposite

Step2: Determine velocity

Velocity is -30 m/s (downward)

(c):

Step1: Displacement s=0, use \(s = ut + \frac{1}{2}at^2\)

\(0 = 30t - 4.9t^2\Rightarrow t(30 - 4.9t)=0\)

Step2: Solve for total time (t≠0)

\(t=\frac{30}{4.9}\approx6.12\ \text{s}\)

Answer:

(a):
≈45.92 m