I cannot get the formula right for this one, can anyone help?

Question

A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m?

Answer 1

The height at time t is

h = 1/2gt^2 + v0*t + h0

h0=initial height=100
v0=initial velocity=20
g=gravity constant=-9.8

Set h=80 and you get a quadratic equation for t that you can solve using the quadratic formula. You will get two values of t, but one of them will be negative, so take the other one.

Answer 2

Well, it has been a while since physics. The ball will reach a height of 80m (bottom building - 80 m up) after 4.9201734 sec

Initial velocity: 20m/s
final velocity (highest point): 0m/s
acceleration= -9.8 m/s2
time 1=???

vf=vi+at
0=20-9.8t
-20=-9.8t
time 1=2.0408 sec

d1=(time 1)(vf + vi)/2
d1=2.0408(10)
d1=20.408 m

then from its highest point to 80 m from the bottom of the building. Since the building is 100m, then you just need 20 more meters.

d=40.408 m
a=9.8 m/s2
vi=0 m/s (it used to be final velocity before)
vf=???

vf2= vi2 + 2ad
vf2= 0 + 2(9.8)(40.408)
vf=28.1424 m/s

then time 2=????

a=(vf - vi)/t
at=28.1424
t=28.1424/9.8
time 2= 2.8716 sec

time 1 + time 2
so the total time is 4.9201734 sec

and that's when the ball reaches 80 m high

IF I AM WRONG LET ME KNOW

Answer 3

I don't want to solve it but maybe this way of thinking will help. After it reaches its peak and then comes back down, isn't it true that when it's back where it started, 100 meters, it's travelling the same speed but down rather than up?

gt=20 m/s should give the travel time to the peak. Twice this is the round trip time back to 100 meters. Then gt^2+20t = 20 meters should give the time to travel from 100 down to 80. Solve for both t values and add.

Answer 4

S=?
U=20m/s
V=0
F=-9m/s/s
Using V*V=U*U-2FS
gives a maximum height of 400/2*9=22.2m above top of building
then
S=22.2+20 m (to fall to an altitude of 80m)
U=0
V=?
F=9m/s/s
T=?
Using S=UT+1/2FT*T
gives T=root of 2*82.2/9=4.2 seconds

Answer 5

Never! It'll hit the roof first on the way down!

HA!

Answer 6

s=ut+1/2at^2 where s is distance, u is initial velocity, a is acceleration (in this case -9.8ms^-2 (gravity)) and t is time

Answer 7

The ball reached 80 m on the way to the roof, (100 m tall building). Duh!!!

Answer 8

h=100-4.9t^2+20t
80=100-4.9t^2+20t
4.9t^2-20t-20=0
t=(20+/-sqrt(400+4*4.9*20))/9.8=(20+/-sqrt(792))/9.8
t=(20+/-28.14)/9.8=4.912 sec.

the other answer is<0 which isn't physically possibly.