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3 years ago

A soccer player kicks a ball horizontally at 25.0 m/s from a bridge, and the ball hits

Physics

1 answer:

san4es73 [151]3 years ago

5 0

Answer: 11.025 meters.

Explanation:

Ok, we know that the initial velocity is only horizontal, so it does not affect the vertical problem.

Looking only at the vertical problem (because we want to know how high is the bridge) we have that:

The acceleration is the gravitational acceleration, g = 9.8m/s^2

A(t) = -9.8m/s^2

Where the negative sign is because this acceleration is downwards.

For the vertical velocity we integrate over time, as we do not have an initial vertical velocity, there is no constant of integration.

V(t) = (-9.8m/s^2)*t

For the position we integrate over time again, here the constant of integration is the initial vertical position, H, that is the height of the bridge.

P(t) = 0.5* (-9.8m/s^2)*t^2 + H.

Now we know that the ball hits the ground 1.5s after it was kicked, then:

p(1.5s) = 0m

With that we can find the value of H.

0 = 0.5* (-9.8m/s^2)*(1.5s)^2 + H.

H = 0.5*(9.8m/s^2)*(1.5s)^2 = 11.025m

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Answer:

# of Snickers bars 2

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3 years ago

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B. opposite charge and smaller mass

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4 years ago

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

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3 years ago

What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el

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