Question

a cannon sitting on top of a 40 m mound shoots a projectile with an initial velocity of 4.47 m/s parallel to ground. how far away did it land?

Answer 1

Given that,

The initial height of the projectile, y₀=40 m

The initial velocity of the projectile, v₀=4.47 m/s

The direction of the initial velocity is parallel to the ground. Therefore the x-component of the initial velocity v₀x=4.47 m/s.

And the y-component of the initial velocity is v₀y=0 m/s

From the equation of the motion we have,

`y=y_0+v_(0y)t+(1)/(2)gt^2`

Where y is the final height of the projectile which is zero as it finally hits the ground. And g =-9.8m/s² is the acceleration due to gravity. And t is the time interval of the flight of the projectile.

On substituting the known values in the above equation,

`\begin{gathered} 0=40+0+(1)/(2)*-9.8* t^2 \\ =40-4.9t^2 \end{gathered}`

On rearranging the above equation and simplifying it,

`\begin{gathered} t=\sqrt{(40)/(4.9)} \\ =2.86\text{ s} \end{gathered}`

The x-component of the velocity remains constant as there is no acceleration in that direction.

The horizontal distance travelled or the range of the projectile can be calculated using the formula,

`R=v_(0x)t`

On substituting the known values in the above equation,

`R=4.47*2.86=12.78\text{ m}`

Therefore the projectile will land 12.78 meters away