Question

A ball is thrown vertically up from the edge of a 150 m building with a veloicty of 10m/s. (A) how high did the ball rise? (b) how much time did it take fort he ball to reach this height ? (C) what is the velocity oft he ball as it strikest he ground ? (D) how long does it take for the ball to reacht he ground from its starting point (e) what is the position and velocity 1 sec and 3 sec after it is thrown ?

Answer 1

v = Vi - 9.81 t
v = 0 at top
so
t = 10/9.81 at top (part B)

Htop = 150 + 10 t - 4.9 t^2 (part A)


now simply drops from Htop to ground

Pe at top = m g Htop
Ke at bottom = (1/2) m v^2
so
v = sqrt (2 g Htop) (part C)

we have original t rising to top
now time falling
v = 0 + g Tfall
we know v from part C
so
Tfall = v/9.81
add that to our answer to part B to get total time in air. That is part D

Part E:
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2

Answer 2

Thank you very much

Answer 3

I don't have to add Tfall with the part b beacuse starting point is 150 and not in the air

Answer 4

That t top in part B is the time to slow down on the way up from the top of the building to the max height. 10/9.81 seconds.
THEN, it starts falling from that top point I called Htop. By the way it will pass the roof on the way down at 10 m/s down :)