it was at the top at time 5.8/2=2.9sec
how high was it? It fell in 2.9s
h= 1/2 g t^2=1/2 g 2.9^2
During a baseball game, a batter hits a pop- up to a fielder 78 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 5.8 s, how high does it rise?
2 answers
It spends 2.9s going up and 2.9s coming down.
Coming down, it falls
height = (g/2)t^2 = (4.9)*2.9^2 = 41.2 meters.
The distance of the fielder is not needed.
Coming down, it falls
height = (g/2)t^2 = (4.9)*2.9^2 = 41.2 meters.
The distance of the fielder is not needed.
1 answer