Standing on the top of a 61 meter high building you throw a ball straight up with an initial speed of 43 m/s. How long to the nearest tenth of a second, does it take to hit the ground?

Question

Standing on the top of a 61 meter high building you throw a ball straight up with an initial speed of 43 m/s.  How long to the nearest tenth of a second, does it take to hit the ground?

anonymous · Answer

S = Vt + .5at^2

(43t + 61) = -43t +.5(9.8)t^2

4.9t^2-86t-61=0

solve for t

theeric · Answer

Hi! I'm just back on quick to see what's going on.

And so I found this!

So, my advice is for you to draw a picture if you haven't already! Then, consider the situation in parts! There's the way up, the way down to the top of the building, and the rest of the way down!

Just remember that the throw upwards is going to be sort of symmetric! That means two things:
1) The time taken to get from the building to the top of the throw will be the same as the time it takes to get back down to the building.
2) When it gets down to the building, it will have the opposite velocity of what it had going up!

theeric · Answer

Or there's that!

theeric · Answer

You can solve that with a certain formula! (starts with a 'Q')

If you do it my way, you need at least two kinematic formulas.

theeric · Answer

take care!