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Usimov [2.4K]
3 years ago
12

I need help on this question:

Mathematics
1 answer:
KonstantinChe [14]3 years ago
6 0

Answer:

ehwrjhq

Step-by-step explanation:pwjhreijwhribhbhrtu4i3r3

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David drops a soccer ball off a building. The building is 75 meters tall.
Nezavi [6.7K]

a) 70.1 m

The ball is moving by uniformly accelerated motion, with constant acceleration g=9.8 m/s^2 (acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation

h(t)=h_0 - \frac{1}{2}gt^2

where h_0 = 75 m is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:

h(1 s)=75 m - \frac{1}{2}(9.8 m/s^2)(1 s)^2=70.1 m

b) 3.9 s

We can still use the same equation we used in the previous part of the problem:

h(t)=h_0 - \frac{1}{2}gt^2

This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have

0=h_0 - \frac{1}{2}gt^2

And solving for t we find

t=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2(75 m)}{9.8 m/s^2}}=3.9 s

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Hope this would help you

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I think the is "A". 50 randomly selected users of the centers facilities.

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APP IT will HELP SIDJIDNUDNUD

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Answer:

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Step-by-step explanation:

I'm in middle school I wish I could help.

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