Connect with a global community of knowledgeable individuals on IDNLearn.com. Our platform offers reliable and detailed answers, ensuring you have the information you need.
Sagot :
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,
[tex]y=y_0+v_{0y}t+\frac{1}{2}gt^2[/tex]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,
[tex]\begin{gathered} 0=40+0+\frac{1}{2}\times-9.8\times t^2 \\ =40-4.9t^2 \end{gathered}[/tex]On rearranging the above equation and simplifying it,
[tex]\begin{gathered} t=\sqrt{\frac{40}{4.9}} \\ =2.86\text{ s} \end{gathered}[/tex]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,
[tex]R=v_{0x}t[/tex]On substituting the known values in the above equation,
[tex]R=4.47\times2.86=12.78\text{ m}[/tex]Therefore the projectile will land 12.78 meters away
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.
