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12345 [234]
2 years ago
7

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

esistance)?
Physics
1 answer:
Brut [27]2 years ago
8 0

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

d=ut-\dfrac{1}{2}gt^2

Putting all values

d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

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Answer:

310 meters

Explanation:

Given:

v₀ = 0 m/s

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a = -9.8 m/s²

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Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²

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Rounded to two significant figures, the object fell 310 meters.

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