Question

A ball is thrown upwards with a speed of 5m/s from a top 20m high building.
a. how long does it take the ball to hit the ground?
b how fast is the ball moving?

Answer 1

a. you need to find time so find a kinematic equation that involves distance, gravitational acceleration, and initial velocity but is missing time

b. you need to find the final velocity so find a kinematic equation that is missing final velocity but includes the aforementioned data

Answer 2

d= -1/2at^2
v(final)= V(inital)-at

use these two equations......

Answer 3

ok thanks

Answer 4

i got a weird number

Answer 5

did you calculate it ? what did ypu get

Answer 6

@oOKawaiiOo ,
\(a=\dfrac{\Delta v}{\Delta t}=\dfrac{v-v_0}{\Delta t}\\\text{(multiply both sides by }t\text{ to get:)}\\\implies a\ \Delta t=v-v_0\\\text{(add }v_0\text{ to both sides to get:)}\\\implies a\ \Delta t+v_0=v\)

Also I believe that your \(d=-\frac{1}{2}at^2\) should come from \(d=d_0+v_0\ t+\frac{1}{2}a\ t^2\).

And I would agree that you need to find a good equation. I mean, you need an equation that has what you want (time to hit the ground) and that has all the other variables being something you know!

You know \(a=g\) in free fall (only gravitational force), \(v_0=5\ [m/s]\), \(d_0=20\ [m]\), and you want \(\Delta t\) (the time interval). You could say \(t_0=0\) and you want \(t\), as an alternative view.

Do you know the equation that is \(v^2=v_0^2+2\ a\ d\)? :) That will get you part B. Then you'll have \(v\) and you can use the definition of acceleration, \(a=\dfrac{\Delta v}{\Delta{t}}\), and solve for \(\Delta t\) for part A.

Unless I can think of an equation to get right to part A!`

Answer 7

i had 4 seconds is that right for part a

Answer 8

Sounds reasonable. I haven't calculated it yet.

Answer 9

It sounds reasonable. It's not what I got, but I could still be wrong. What did you try?

Answer 10

Wait, I know I made a mistake!

Answer 11

I still have a different result...

Answer 12

how did you plug it in

Answer 13

may be im doing something wrong

Answer 14

I did it backwards, but I'll combine the equations...

\(v=-\left|\sqrt{v_0^2+2\ a\ \Delta d}~\right|\)
\(a=\dfrac{v-v_0}{t}\implies v=a\ \Delta t+v_0\)

Substituting \(v=a\ \Delta t+v_0\) into the first equation,
\(v=a\ \Delta t+v_0=-\left|\sqrt{v_0^2+2\ a\ \Delta d}~\right|\\\ \\\ \\\implies\Delta t=\dfrac{-\left|\sqrt{v_0^2+2\ a\ \Delta d}~\right|-v_0}{a}\)

Answer 15

There must be an equation that I forget or never knew to make part A easier...

Answer 16

thanks a lot

Answer 17

You're welcome :)