Question

Question #4f1eb

Answer 1

Explanation:

Explanation:

We know the ball is thrown upwards in a initial velovity of `15.8 m/s`

So, `u=15.8 m/s`

When the ball reaches its maximum height, it has a velocity of `0 m/s`

So, `v=0 m/s`

Let the maximum height be `x`

The gravitational force that makes the ball slow down till it reaches its maximum height

So, `a=-9.8 m/s`

Now, we have to find the time to reach its maximum height

We have our values

`color(orange)(u=15.8 m/s`
`color(orange)(v=0 m/s`
`color(orange)(a=-9.8 m/s`

We can use one of the Kinematic equations

`color(blue)(v=u+at`

`rarr0=15.8+(-9.8t)`

`rarr0=15.8-9.8t`

`rarr9.8t=15.8`

`rarrt=15.8/9.8`

`color(green)(rArrt=1.61 s`

We can use this value to find the maximum height (`x`) using another Kinematic equation

`color(brown)(x=ut+1/2at^2`

`rarrx=(0*1.61)+(1/2*(-9.8)*1.61^2)`

`rarrx=1/2*(-9.8)*2.59`

`rarrx=(-9.8)*1.29`

`color(green)(rArrx=-12.64m`

As height cannot be negative, `x=12.64m`

Hope this helps! :)