At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
you just confused me.
My bad. Are you familiar with calculus, namely derivatives?
OK. No problem. I can help you solve it using another method. Stand by
\[-4.9t^2+49t=0\]solve for \(t\)
h = −4.9t2 + vt + s First let's try to understand this formula and what it means. h is telling you how high the projectile is at ANY point in time. -4.9t^2 is telling you that gravity is acting downward. vt is allowing you to specify the initial velocity (they gave you this information, namely 4.9 meters per second). s is telling you where the projectile is starting (they also told you this, namely at the ground or ZERO). How are you doing so far?
ok. i got that.
satellite73 gave you the short and sweet version. Let me know if you know how to solve a quadratic.
yea i do.
i got t=o,-10
Satellite also gave you the wrong formula. He wrote 49t. That means the ball is moving initially at 49 meters per second. Is that what the problem gave you?
t=0 is a correct solution. That is telling you that the ball was at a height of ZERO, the clock started. Your second solution is the answer you need.
Solve 0=-4.9t^2+4.9t You should get t=0 t=.9574 (not rounded)
ohhh. yay! thanks!