Answer:
1.62 m/s
Explanation:
Wavelength of the water wave= 54 m
The frequency is 0.03 Hz
Therefore the velocity can be calculated as follows
Velocity= frequency × wavelength
= 0.03 × 54
= 1.62 m/s
Acc to
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<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>
<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>
<span>t = √(2y/g) </span>
<span>in the ft - lb - s system </span>
<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>
<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation

where v is final velocity which is zero at max height and u is it initial
hence


now we can find time in the 15 cm ascent


using quadratic formula

t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using

where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom


when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is

t=0.0409