initial vertical velocity:
at the top, vf in vertical, is zero. so in the vertical,
vf^2=vi^2+2ad
0=vi^2-2*9.8*10
vi=sqrt(2*9.8*10)
so v=sqrt(vi^2 + (1/2 vi)^2)
v=2*9.8*10 sqrt(1+1/4)
= 2*9.8*10/2 sqrt 5
time in air:
vf=vi+at
0=sqrt(20*9.8)-9.8t solve for t
then knowing t,
horizontal distance=vhor*timeinai
= 1/2 (sqrt(20*9.8))*98 sqrt5
Searches related to A football at rest is kicked so that it start to move with a velocity where horizontal component is 1/2 vm/s. In it's flight , the ball rises to a maximum height of 10m.Assuming that air resistance may be neglected and that the ground is horizontal , calculate : i) Value of v. (ii) horizontal distance travelled by the ball before striking the ground , take g =10m/s
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