Posted by **Anonymous** on Saturday, August 25, 2012 at 8:30pm.

Standing on the top ledge of a 55 m high building you throw a ball straight up with an initial speed of 45 m/s. How long to the nearest second, does it take to hit the ground?

I got 10 seconds for the first question.

How do I go about answering this one?

If the building in the previous problem is 48 meters high and you throw the ball up at 30 m/s, how high to the nearest meter does it go?

- Physics -
**Damon**, Saturday, August 25, 2012 at 8:43pm
well, once it stops at the top, it is a simple fall from there.

How high above the building does it go and how long to the top?

(45 m/s is quite a throw straight up)

m g h = (1/2) m v^2

h = (1/2) (45)^2/9.81 = 103 m

average speed up = 45/2 = 22.5 m/s

so time in air up =103/22.5 = 4.58 s upward

now it falls from 55+103 = 158 m

d = (1/2) g t^2

158 = 4.9 t^2

t = 5.68 s

total time in air = 4.58 + 5.68 = 10.3 s

the way I did it the second question is really easy

- Physics -
**Anonymous**, Sunday, August 26, 2012 at 12:30am
The correct answer was 94 meters for the second part of the question. That's what I'm unsure as to how to get it...

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