Respuesta :
For this problem, we can use a kinematic formula to help us solve this. We know the given the distance, the initial velocity, and the acceleration due to gravity, which is 9.8 m/s². We want to know the final velocity.
[tex]v_f^2=v_i^2+2ad[/tex]
[tex]v_f[/tex] is the final velocity
[tex]v_i[/tex] is the initial velocity
[tex]a[/tex] is the acceleration, in this case, the acceleration due to gravity
[tex]d[/tex] is the distance.
[tex]v_i=0 \frac{m}{s}[/tex]
[tex]d=13m[/tex]
[tex]a=9.8 \frac{m}{s^2}[/tex]
[tex]v_f=??? \frac{m}{s}[/tex]
[tex]v_f^2=(0 \frac{m}{s})^2+2(9.8 \frac{m}{s^2})(13m)[/tex]
[tex]v_f^2=254.8 \frac{m^2}{s^2} [/tex]
Take the square root of both sides.
[tex]v_f \approx 15.96 \frac{m}{s} [/tex]
That is the velocity right before the shell hits the rocks. I hope this helps! :)
[tex]v_f^2=v_i^2+2ad[/tex]
[tex]v_f[/tex] is the final velocity
[tex]v_i[/tex] is the initial velocity
[tex]a[/tex] is the acceleration, in this case, the acceleration due to gravity
[tex]d[/tex] is the distance.
[tex]v_i=0 \frac{m}{s}[/tex]
[tex]d=13m[/tex]
[tex]a=9.8 \frac{m}{s^2}[/tex]
[tex]v_f=??? \frac{m}{s}[/tex]
[tex]v_f^2=(0 \frac{m}{s})^2+2(9.8 \frac{m}{s^2})(13m)[/tex]
[tex]v_f^2=254.8 \frac{m^2}{s^2} [/tex]
Take the square root of both sides.
[tex]v_f \approx 15.96 \frac{m}{s} [/tex]
That is the velocity right before the shell hits the rocks. I hope this helps! :)
