First note that the velocity is always constant in the x direction, meaning that acceleration in the x direction would be equal to 0.
So : dfx = 0.5*ax* (t^2) + v0x*t + d0x ax=0 vox=27.9 m/s d0 = 0 dfx = 19.9 m
so it becomes: dfx = v0x * t => 19.9 = 27.9 * t => t= 19.9 / 27.9 = 0.713 s
So the total time would be 0.713 s.
Now we can jump into y-direction. We already figured that the total time is 0.713 s. This is also equal to the amount of time the ball travels vertically, until it hits the ground.
So: dfy =0.5*ay* (t^2) + v0y*t + d0y where ay=g= -9.8 m/s^2 v0y= 0 m/s dfy = 0 m d0y=?
Please note that by writing v0y what I mean is the initial velocity in the y direction, and by writing dfy I mean the final position in the y direction (which is zero since the ball hits the ground, so vertical position relative to ground is 0).
So we have: 0 = 0.5* (-9.8) * (0.713^2) +( 0 * (0.713)) + d0y
so d0y = 0.5* (9.8) * (0.713^2) = 2.491 m
Please let me know if you have any questions.
