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ddd [48]
3 years ago
15

A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later

. How
high is the cliff and how far from the base of the cliff did the diver hit the water?
Help asap
Physics
1 answer:
mart [117]3 years ago
8 0

Explanation:

It is given that,

The horizontal speed of a cliff diver, v_x=3.6\ m/s

It reaches the water below 2.00 s later, t = 2 s

Let d_x is the distance where the diver hit the water. It can be calculated as follows :

d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m

Let d_y is the height of the cliff. It can be calculated using second equation of motion as follows :

d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

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How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have
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1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
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What is a asteroid traveling rapidly called​
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Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp
Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

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