Question

Question #04fd1

Answer 1

The ball will reach a max height of `ul45` m in `ul3` seconds

Explanation:

We use the kinematic equation connecting various quantities of interest. The maximum height is reached when upwards velocity is `=0`
1. To calculate Max height `h` the kinematic equation is
`v^2-u^2=2ah`
Final velocity `v` after time `t`, `a` is acceleration and initial velocity is `u`

Inserting given values we get, remember, direction of acceleration due to gravity is opposite to the direction of initial velocity

`0^2-30^2=2xx(-10)h`
`=>20h=900`
`=>h=900/20=45m`
2. To calculate time taken to reach maximum height we use the other kinematic equation
`v=u+at`

Inserting given values we get

`0=30-10xxt`
`=>10xxt=30`
`=>t=30/10=3s`

Answer 2

h= 45 m; t = 3 s

Explanation:

The initial Kinetic energy (where potential energy is zero) becomes only potential energy (where the ball reaches the max height and its speed is zero).
`1/2 m v^2 = mgh`
but, since the mass is the same
`1/2 v^2 = gh`
and
`h= v^2/(2g)`
`h = (900 m^2/s^2)/(2 10 m/s^2)= 45 m`
from the second law of motion
`v= v°-at`
`0= 30 m/s-10m/s^2 t`
`t= 3 s`
now let's verify in the first law of motion
`s= v°t -1/2 g t^2`
`45m = 30 m/s 3s - 1/2 10 m/s^2 (3s)^2`
45 = 45 m