Asked by Anonymous
During a baseball game, a batter hits a high pop-up. if the ball remains in the air for 4.2 seconds, how high does it rise? The accelleration of gravity is 9.8 m/s^2.
Answers
Answered by
Scott
ignoring the height of contact
time up equals time down
h = .5 g t^2 = .5 * 9.8 * 2.1^2
time up equals time down
h = .5 g t^2 = .5 * 9.8 * 2.1^2
Answered by
some highschool kid ya kno
idk what he wrote so here ya go
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
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
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