Having trouble with a simple 1-D physics problem... help. :(?

Question

A ball is thrown vertically upwards with an initial velocity of 20.08 m/s. How long is the ball in the air? t= 4.09 s

What is the greatest height reache dby the ball? x= 20.6 m

**** Calculate the time at which the ascending ball reached a height of 15 m. ???

I'm stuck on the last problem. I tried plugging it into a qudratic equation 15m=0+20.08t + 1/2(9.8)t^2 so 4.9t^2 + 20.08t - 15. The answer was 0.64 s but it didnt' work and doesn't make since. I also trie comparing the two numbers so [2.049 s (the time it takes to reach the maximum height) * 15m/20.57m) = 1.5 s. That made more since to me, but this also didn't work. Any hints?

I'm really tired, excuse my poor grammar and typos.

Answer 1

Sign problem as written.

Gravity is down, velocity is up

15 = -4.9t^2+20.08t


on the way up: 0.9826 s
on the way back down: 3.1153 s

(I did this with my TI-83, not by solving the quadratic.)

Answer 2

Calculate the time at which the ascending ball reached a height of 15 m.

"...I'm stuck on the last problem. I tried plugging it into a qudratic equation 15m=0+20.08t + 1/2(9.8)t^2..."

your equation is slightly incorrect. Gravity points down, so we give it a negative sign:

15m=20.08t + 1/2( -9.8 )t^2

when you solve this you'll get:

t1 = 3.1153
and
t1 = 0.9826

so, on the way up the ball is 15m high at t = 0.9826s, and on the way down it is again at 15m at t =3.1153.

Answer 3

i hope that you need not use equations that may really confuse you.
the simple way to solve this is trying to imagine how the ball behaves during the travel, and of course the formula v=d/t applies.
you see, total height is equal to the initial velocity multiplied by the time it took there (vo to) ;
plus the second velocity rate and the time it took unitl it reach a total stop because of gravity pull (v1 t1)

the initial time would be to = 4.09 -t1, where explained is the total time (4.09) minus the time of the second velocity.

thus we have the equation

h = vo to + v1 t 1 (putting all value we have would give us)
20.6 = 20.08(4.09-t1) + v1 t1

we then use v1 t1 = 15

20.6 = 82.13 - 20.08t1 + 15
20.08t1 = 82.13 + 15 - 20.6
t1 = 76.53/20.08

thus we have the answer:

t1 = 3.8 secs


keep on studying ! enjoy !

Answer 4

I'm getting t=.645 when I do it, perhaps you just made a little mistake somewhere in the quadratic formula? You're using the right equation and stuff :)