Question

Question #68f57

Answer 1

1) Velocity after 2 second
`v=10.4 m/s`

2) Maximum Height Attained
`s=45.91 m`

3) Time Taken to reach the Maximum height
`t=3.06 sec`

4) Time taken to the ball to return to the ground
`T=6.06 sec`

Explanation:

Initial Velocity `u=30 m/s`
For upward motion Acceleration due to gravity `a=-g=-9.8 m/s^2`
1) Velocity after 2 second
`t=2 sec`
Using Newton's First Equation of Motion
`v=u+at`
`v` is the final velocity after t seconds.
`v=30-g t=30-(9.8)(2)=30-19.6=10.4 m/s`
`v=10.4 m/s`

2) Maximum Height Attained
On the Maximum height final velocity of object will be zero.
`v=0`
Using Newton's Third Equation of Motion
`v^2=u^2+2as`
s is distance traveled
`=>0=30^2-2(9.8)s`
`19.6s=900`
`s=900/19.6=45.91 m`

3) Time Taken to reach the Maximum height
On the maximum height final Velocity `v=0`
Using Newton's First Equation of Motion
`v=u+at`
`v` is the final velocity after t seconds.
`0=30-g t=30-(9.8)(t)`
`9.8t=30`
`t=30/9.8`
`t=3.06 sec`

4) Time taken to the ball to return to the ground
The time taken to reach maximum height and return to the ground is the same.
Hence Total time `=2xx`time taken to reach the maximum height
`T=2xx3.06=6.06 sec`