Question

How fast must a ball be thrown vertically from the bottom up to reach a height of 50m? How long it will remain in the air?

Answer 1

remeber the kinematics, you will need two of the equations for this one:
first: vf^2=vi^2+a*d
then once you find vi, plug it into the second:
vf=vi+a*t
when throwing a ball up, when it reaches its max height, the final velocity is 0. and the distance we are looking for is 50, and acceleration due to gravity is -9.8m/s^2
so:
0=vi^2+(-9.8m/s^2)*50m
so -vi^2=-490 m^2/s^2
vi=sqrt(490m^2/s^2)
vi=22.14m/s
then plug it in:
vf=vi+a*t
0=22.14m/s+(-9.8m/s^2)*t
-22.14=-9.8*t
-22.14m/s/-9.8m/s^2=t
t=2.26s

** don't hate me if it's wrong :(( **

Answer 2

um...i think ur formula is wrong
im pretty sure it is
vf^2=vo^2+2ad

Answer 3

otherwise i think ur thought process is correct

Answer 4

i got initial speed 31.32m/s and time 6.38s

Answer 5

Perfect Cherio12, i got this result now tbem

Answer 6

no problem