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

Step-by-step 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=(u)/(g)\\\\t=(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-(1)/(2)gt^2`

Putting all values

`d=30(3.06)-(1)/(2)* 9.8* (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.