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During a baseball game, a batter hits a high pop up. If the ball remains in the air for 4.16 S, how high above the point where...Asked by isa
During a baseball game, a batter hits a high pop-up.
If the ball remains in the air for 6.41 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat.
acceleration of gravity is 9.8 m/s
If the ball remains in the air for 6.41 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat.
acceleration of gravity is 9.8 m/s
Answers
Answered by
isa
During a baseball game, a batter hits a high pop-up.
If the ball remains in the air for 6.41 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat.
acceleration of gravity is 9.8 m/s
can someones pls help im currently doing the timed homework
If the ball remains in the air for 6.41 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat.
acceleration of gravity is 9.8 m/s
can someones pls help im currently doing the timed homework
Answered by
R_scott
time up equals time down
d = 1/2 * g * (6.41 / 2)^2
your g number is two sig fig ... so your answer should be also
d = 1/2 * g * (6.41 / 2)^2
your g number is two sig fig ... so your answer should be also
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