Answer:
a) Answer: 1.428 seconds.
The formula for free fall from height x is:
x (m) = 1/2 • g (m/s^2) • t^2 (s^2)
The time t (s) for free fall from height x (m) is:
t (s) = SQR[2 • x (m) / g (m/s^2)]
So when height x = 10 m,
t (s) = SQR[2 • 10 (m) / 9.8 (m/s^2)]
t (s) = SQR[20 (m) / 9.8 (m/s^2)]
t (s) = SQR[2.0408 s^2]
t (s) = 1.428 (s).
What if t = 1.00 (s)? How high is that?
x (m) = 1/2 • 9.8 (m/s^2) • 1.00^2 (s^2).
x = 4.9 m (16.07 ft).
Consequently, any planet’s gravity acceleration g can be found simply by dropping balls from various heights until the drop height results in t = 1.0 (s), and then multiply that height by 2.
On the Moon for example, g is officially 1.625 m/s^2. The 1.0 s drop height is half g, so 0.8125 m. Once you find the drop height at t = 1.0 s, the moon’s g is twice that
b) Answer:
given: d=100m g=9.8m/s² Vi=0
unknown; Vf=? t=?
formula: Vf²=Vi²+2gd
t=vf-vi/g
solution:Vf²=0²+2×9.8m/s²×100m
t=44.27m/s - 0m/s/9.8m/s²
√Vf²=√1960 m²/s²
answer; Vf=44.27m/s
t=4.52s
