during a baseball game, a batter hits a pop-up to a fielder 88 m away. the acceleration of gravity is 9.8m/s/s. if the ball remains in the air for 6.1seconds, how high does it rise? answer in units of meters.
2007-11-12 04:31:19 · 2 answers · asked by broncosfanatic09 1 in Science & Mathematics ➔ Physics
the ball rises to it's max height for half of the flight = 6.1/2 = 3.05 sec,
and falls back to the round for the other half = 3.05 sec
An object falling for 3.05 sec covers a distance equal to .5gt^2 so the height was:
height = .5 (9.8) (3.05)^2 = 45.6 meters
2007-11-12 04:44:59 · answer #1 · answered by ignoramus_the_great 7 · 0⤊ 0⤋
The horizontal velocity is vh= 88/6.1= 14.43 m/s
h=-4.9 t^2 +v(vertical)*t
h=0 for t=0 and for t= v(vertical)/4.9= 6.1
so v(vertical= 29.89m/s
so we can write
h= -4.9 t^2+29.89 t
h´= -9.8 t +29.89= 0 so t = 29.89/9.8 = 3.05 s (for h max)
h max = 45.58m
2007-11-12 12:50:24 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋