Answer:
3.1 m/s
Explanation:
First, find the time it takes for the cat to land. Take down to be positive.
Given:
Δy = 0.61 m
v₀ = 0 m/s
a = 9.81 m/s²
Find: t
Δy = v₀ t + ½ at²
(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²
t = 0.353 s
Now find the horizontal velocity needed to travel 1.1 m in that time.
Given:
Δx = 1.1 m
a = 0 m/s²
t = 0.353 s
Find: v₀
Δx = v₀ t + ½ at²
(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²
v₀ = 3.1 m/s
Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?
Answer:
measure the position every so often with a stopwatch
Explanation:
A possible method of measurement is to place a measuring tape along the path and measure the position every so often with a stopwatch, with this we can make a graph of position against time and by extrapolation find the initial velocity.
This is a method used in measurements of uniform movements of bodies
solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
