Question

A firefighting crew uses a water cannon that shoots water at 30.0 m/s at a fixed angle of 45.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1

Answer 1

The distance from the building is 11.41 m.

Step-by-step explanation:

Given that,

Speed = 30.0 m/s

Angle = 45.0°

Distance = 10.0 m

We need to calculate the vertical component

Using formula of vertical component

`u_(y)=u\sin\theta`

`u_(y)=30.0*\sin45`

`u_(y)=21.21\ m/s`

We need to calculate the time

Using equation of motion

`s=ut-(1)/(2)gt^2`

Put the value into the formula

`10=21.21t-(1)/(2)*9.8* t^2`

`21.21t-(1)/(2)*9.8* t^2-10=0`

`-4.9t^2+21.21t-10=0`

`t=0.538\ sec`

On neglecting of higher value of t

Then use the horizontal component

`u_(y)=u\cos\theta`

`u_(x)=30.0*\cos45`

`u_(x)=21.21\ m/s`

We need to calculate the distance

`d=v* t`

Put the value into the formula

`d=21.21*0.538`

`d=11.41\ m`

Hence, The distance from the building is 11.41 m.