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
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?
2 answers
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...
1 answer