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4 years ago

Please helpppppppppp

Physics

1 answer:

Readme [11.4K]4 years ago

6 0

The height of the cliff is 96.2 m

Explanation:

The motion of the rock along the horizontal direction is a uniform motion (constant velocity), since there are no forces acting along that direction. Therefore, the horizontal velocity is given by

$v_x = \frac{d}{t}$

where, using the data of the first rock:

d = 66.5 m is the horizontal distance travelled by the rock

$v_x = 15 m/s$ is the horizontal velocity

Solving for t, we find the time it takes for the rock to reach the ground:

$t=\frac{d}{v_x}=\frac{66.5}{15}=4.43 s$

The vertical motion instead is a free fall motion acted upon gravity, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the vertical displacement of the rock, which corresponds to the height of the cliff

u = 0 is the initial vertical velocity of the rock

t = 4.43 s is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity (we chose downward as positive direction)

Solving for s, we find the height of the cliff:

$s=0+\frac{1}{2}(9.8)(4.43)^2=96.2 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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Answer:

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Explanation:

Given data

$v_{o}=7.14m/s\\\alpha =36.5^o$

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$a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2$

Using equation of motion

$v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m$

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Using the result in Part (a) we can substitute in other equation of motion to get time t:

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At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

$a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s$

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