Question
A beanbag was thrown straight upward from the ground at a initial velocity of 22.5 m/s. At the same time, another beanbag is dropped from the top of 17. 0 m building. A what time will the two beanbags be at the same height above the ground?
Answers
bobpursley
going upward:
hf=22.5*t-4.9t^2
going downward:
hf=17-4.9t^2
but hf=hf, and times are same, so
17-4.9t^2=22.5*t-4.9t^2
or t=17/22.5 seconds
hf=17-4.9(17/22.5)^2=14.2 ? meters
hf=22.5*t-4.9t^2
going downward:
hf=17-4.9t^2
but hf=hf, and times are same, so
17-4.9t^2=22.5*t-4.9t^2
or t=17/22.5 seconds
hf=17-4.9(17/22.5)^2=14.2 ? meters