Question

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 resistance)?

Answer 1

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.