Asked by arturo bailon
During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 6.65 s, how
high does it rise? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m
CAN ANYONE HELP PLEASE ??????!!!
pop-up.
If the ball remains in the air for 6.65 s, how
high does it rise? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m
CAN ANYONE HELP PLEASE ??????!!!
Answers
Answered by
Damon
falls from top for 6.65/2 = 3.325 seconds
h = (1/2) (9.8) (3.325)^2
h = (1/2) (9.8) (3.325)^2
Answered by
Anonymous
can u help me pleaseeeeee @Damon
Answered by
some highschool kid ya kno
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
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