Asked by ruchiket
Two stones are thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with 3 seconds in between, what is the height at which they collide?
Answers
Answered by
Steve
1st: 49t-4.9t^2
2nd: 49(t-3)-4.9(t-3)^2
So, when are the heights equal?
49t-4.9t^2 = 49(t-3)-4.9(t-3)^2
t=6.5
Now just plug that in to find the height.
2nd: 49(t-3)-4.9(t-3)^2
So, when are the heights equal?
49t-4.9t^2 = 49(t-3)-4.9(t-3)^2
t=6.5
Now just plug that in to find the height.