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goldfiish [28.3K]
1 year ago
15

For general projectile motion, the horizontal component of a projectile's velocity aSelect one:O a. continuously increases.O b.

continuously decreases.O c. is zeroO d. remains a non-zero constant.O e. any of the above, depending on position.
Physics
1 answer:
liubo4ka [24]1 year ago
3 0

ANSWER:

d. remains a non-zero constant.

STEP-BY-STEP EXPLANATION:

If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant

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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

hence

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

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now we can find time in the 15 cm ascent

s=ut+0.5at^2

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using quadratic formula

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

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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}

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