No way it actually falls with one g while approaching terminal velocity (There is an increasing force up) but I guess we will have to assume that.
Inititial phase
v = - 9.81 t
h = 560 - 4.9 t^2
so when t = 8
v = -9.81*8 = - 78.5 m/s (terminal velocity)
h = 560 - 4.9*64 = 560-314 = 246 meters high (so fell 314 meters so far)
Now part 2
initial height = Hi = 246
constant speed = Vi = -78.5
so 246 meters /78.5 m/s = 3.13 seconds more
8 + 3.13 = 11.1 seconds total
A ball is dropped from rest at the top of a 560 meter tall building. The ball freely falls with constant acceleration for 8 seconds; at this time, the ball reaches a terminal velocity.
a)How far has the ball fallen when it reaches terminal velocity?
b) What is the terminal velocity of the ball?
c)What is the total time that the ball takes to reach the ground?
1 answer
1 answer