Tf = T/2 = 6.37/2 = 3.19 s. = Fall time.
h = 0.5*g*Tf^2 = 4.9*3.19^2 =
During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 6.37 s, how
high does it rise? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m
2 answers
vf-vi=at, so find that negative number then divide by two and it will give u |vi|
make it positive
vf =0 at the top, so
0^2=vi^2 + 2a ∆x
then
-(vi^2)/(2a){should be 19.6}= ∆x
make it positive
vf =0 at the top, so
0^2=vi^2 + 2a ∆x
then
-(vi^2)/(2a){should be 19.6}= ∆x