Question

a student drops a ball from a height of 54.0m. if the ball remains in the air for 3.29s, determine all unknowns and answer the following question. assume the balls speed changes steadily.
si = 54 m
s = 16.4 m
sf = 0 m
d = 54 m/s
t = 3.29 s
what was the balls speed just before striking the ground?
32.8 m/s

Explanation:

Step1: Identify the motion type

The ball is in free - fall (uniformly accelerated motion) with initial velocity \(u = 0\ m/s\) (dropped), displacement \(s=54.0\ m\), time \(t = 3.29\ s\). We can use the kinematic equation \(v=u + at\) or \(s=ut+\frac{1}{2}at^{2}\) or \(v^{2}=u^{2}+2as\). Another way is to use the average velocity formula. The average velocity \(\bar{v}=\frac{s}{t}\), and for uniformly accelerated motion \(\bar{v}=\frac{u + v}{2}\) (where \(v\) is the final velocity).

Step2: Calculate average velocity

The displacement \(s = 54.0\ m\) and time \(t=3.29\ s\). The average velocity \(\bar{v}=\frac{s}{t}=\frac{54.0}{3.29}\approx16.4\ m/s\) (this matches the given value of \(s\) in the problem's interface, probably a typo in the variable name, it should be average velocity or something related).

Step3: Use average velocity formula for final velocity

Since \(u = 0\ m/s\) (dropped) and \(\bar{v}=\frac{u + v}{2}\), we can solve for \(v\). Rearranging the formula \(\bar{v}=\frac{0 + v}{2}\), we get \(v = 2\bar{v}\). We know \(\bar{v}=\frac{54.0}{3.29}\approx16.4\ m/s\), so \(v=2\times16.4 = 32.8\ m/s\) (or using \(v=u+at\), first find \(a\) from \(s = ut+\frac{1}{2}at^{2}\), with \(u = 0\), \(a=\frac{2s}{t^{2}}=\frac{2\times54.0}{(3.29)^{2}}\approx\frac{108}{10.8241}\approx9.98\ m/s^{2}\approx10\ m/s^{2}\), then \(v=u + at=0+9.98\times3.29\approx32.8\ m/s\)).

Answer:

The ball's speed just before striking the ground is \(\boldsymbol{32.8\ m/s}\)