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marin [14]
4 years ago
8

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top

of the cliff. the height of the cliff is 5.39 m. the stones are thrown with the same speed of 9.44 m/s. find the location (above the base of the cliff) of the point where the stones cross paths.
Physics
1 answer:
Sloan [31]4 years ago
5 0
<span>The sum of the speeds is 9.44 m/s + 9.44 m/s which is 18.88 m/s Note that the sum of the speeds of the two stones is always 18.88 m/s because as the upward moving stone loses speed, the downward moving stone gains the same amount of speed each unit of time. We can find the time for the stones to meet. t = d / v t = 5.39 m / 18.88 m/s t = 0.285487 seconds We can use the upward moving stone to find the height y. y = v0 t + (1/2) a t^2 y = (9.44 m/s)(0.285487 s) - (1/2) (9.8 m/s^2) (0.285487 s)^2 y = 2.30 m The two stones cross paths a height of 2.30 meters above the base of the cliff.</span>
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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
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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

v^2=u^2+2as

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

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u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

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

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

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=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
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