posted by courtney on .
A ball is thrown vertically upwards with an initial velocity of 20.96 m/s. Neglecting air resistance, how long is the ball in the air? What is the greatest height reached by the ball?
this problem seems easy enough but I can't seem to get the answer. How can I get time using only velocity?
Two equations for constant acceleration:
v = Vo + a t
x = Xo + Vo t +(1/2) a t^2
a = -9.8 m/s^2
Vo = 20.96
Xo = 0
when v = 0, we are at the top
solve for t at the top (t at the ground again will be twice that by the way)
Then use that t at the top in the second equation
x = 0 + 20.96 t -4.9 t^2
to get the height at the top
if you did not belive me about doubling that t at the top to get t when it hits the ground again, there are two ways to proceed.
let the ball drop with zero initial velocity from the top (you will see why the time falling is the same as the time rising)
solve the second equation for t when x= 0,
of course t = 0 and t = something which I hope is t = twice t at the top should be solutions to the quadratic.