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
[tex]t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s[/tex]
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.
[tex]d=ut-\dfrac{1}{2}gt^2[/tex]
Putting all values
[tex]d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m[/tex]
Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.
