A golf ball is putted towards te cup which is located 5.00m away, delectation is 0.250m/s^2, what is the velocity must the putt be made so that the ball just drops into the cup?
so I'm having trouble understanding the question,
so t=?, vi=?, a= -.250m/s^2, si = 0.00m, sf = 5.00m and vf=0m/s? so to do this i did vi= √vf-(2ad) = 1.581m/s but im not sure if that right
The motion of the golf ball in this problem is one-dimensional and with constant acceleration, so we can use a kinematic equation that doesn't involve time $t$ as that value is unknown here. $$\left(v_f\right)^2=\left(v_i\right)^2+2a\left(x_f-x_i\right)\implies v_i=\sqrt{\left(v_f\right)^2-2a\left(x_f-x_i\right)}$$ We know that $x_i=0\;\mathrm{m}$, $x_f=5\;\mathrm{m}$, $a=-0.25\;\left.\mathrm{m}\middle/\mathrm{s}^2\right.$, and $v_f=0\;\left.\mathrm{m}\middle/\mathrm{s}\right.$ is implied. We want to solve for $v_i$ in this problem. \begin{align} v_i&=\sqrt{\left(0\;\left.\mathrm{m}\middle/\mathrm{s}\right.\right)^2-2\;(-0.25\;\left.\mathrm{m}\middle/\mathrm{s}^2\right.)\left(5\;\mathrm{m}-0\;\mathrm{m}\right)}\\v_i&=\sqrt{2.5\;\left.\mathrm{m}^2\middle/\mathrm{s}^2\right.}\\v_i&\approx1.581\;\left.\mathrm{m}\middle/\mathrm{s}\right. \end{align} This is the same result that you found. We can verify our result by using it in the same kinematic equation, but this time calculating a value that we already know beforehand with complete certainty, such as $a$. $$\left(v_f\right)^2=\left(v_i\right)^2+2a\left(x_f-x_i\right)\implies a=\frac{\left(v_f\right)^2-\left(v_i\right)^2}{2\left(x_f-x_i\right)}$$ Let us evaluate $a$ and see if it is equal to the value that the problem gives. $$a=\frac{\left(0\;\left.\mathrm{m}\middle/\mathrm{s}\right.\right)^2-\left(\sqrt{2.5\;\left.\mathrm{m}^2\middle/\mathrm{s}^2\right.}\right)^2}{2\left(5\;\mathrm{m}-0\;\mathrm{m}\right)}=\frac{-2.5\;\left.\mathrm{m}^2\middle/\mathrm{s}^2\right.}{10\;\mathrm{m}}=-0.25\;\left.\mathrm{m}\middle/\mathrm{s}^2\right.$$ This verifies our result. You solved the problem correctly—good job.
