A ball is shot straight up into the air with initial velocity of 50ft/sec. Assuming that the air resistance?

0 ⤊

can be ignored, how high does it go?

2007-09-18 09:25:22 · 3 answers · asked by simonkf2002 1 in Science & Mathematics ➔ Mathematics

Vf^2 = 2ad + Vi^2

when it reaches its maximune height, it speed is 0ft/s

0^2 = 2(-32)d + 50^2
2500 = -64d
d = 39.0625 ft

2007-09-18 09:33:14 · answer #1 · answered by 7 · 0⤊ 0⤋

50t-32 t2=0

I believe it's goes up 1.5625 seconds and reaches a hieght of (1.5625)^2 x 16 = about 39.0625 feet

2007-09-18 16:34:45 · answer #2 · answered by Will 4 · 0⤊ 0⤋

The ball goes up 11.83 meters

2007-09-18 16:30:46 · answer #3 · answered by peteryoung144 6 · 0⤊ 0⤋