Objects falling.

Question

A small ball is released from rest and falls on a horizontal platform which is descending vertically at a constant speed of 7m/s, If the ball is 14 m above the platform at the instant of release, clculate the time that elapses before the ball hits the platform.

Guys here is what I tried: s=distance so $s_{ball}+14=s_{platform} \Rightarrow 4.9t^2+14=7t$, but when I put my equation in the calculator I get an error because there is a square root of a negative number so I must have went wrong somewhere in the equation. Plz someone try it and tell me what answer they get.

Emilio Novati · Accepted Answer · 2015-02-10T21:59:51.343

1

You have signs errors in your equations of motion. If the vertical axis is directed upward the the velocity of the platforma and the acceleration of the ball ar directed downward. So the equation you have to solve is: $$ -\dfrac{1}{2}gt^2+14=-7t $$

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edited Feb 10 '15 at 21:59

answered Feb 10 '15 at 20:11

Emilio Novati

62,675

guys can u please explain why it doesnt work if I find the time it takes the ball to travel 14 m then make an equation in terms of t as follows: 4.9t^2= 7(t+sqrt(20/7)) – Philip O'Neill Feb 10 '15 at 20:42
from where comes $\sqrt{20/7}$?? – Emilio Novati Feb 10 '15 at 20:49
Anyway, the core question is that acceleration and velocity are opposite to the initial displacement of $14$m. – Emilio Novati Feb 10 '15 at 20:52
I added a figure to my answer. Ihope it's useful. – Emilio Novati Feb 10 '15 at 21:29
Thank u very much, u were great help – Philip O'Neill Feb 10 '15 at 21:33
welcome! Don't forget to accept if you like! – Emilio Novati Feb 10 '15 at 21:51

qmd · Answer 2 · 2015-02-10T20:23:59.017

Since they are both falling you can set up the equations for distance and then set them equal:

Distance of the ball: $B(t)=$$\frac{1}{2}gt^2-14m$

Distance of the platform: $P(t)=(7\frac{m}{s})\cdot t$

You want to find the time $t$ when they have travelled the exact same distance.

$\implies B(t)=P(t) \iff \frac{1}{2}gt^2-14m=7\frac{m}{s}\cdot t$

$\iff gt^2-14\frac{m}{s}\cdot t-28m=0 \iff 9.81\frac{m}{s^2}\cdot t^2-14\frac{m}{s}\cdot t-28=0$

I solved this quadratic in mathematica and I got

$t=2.55\space s$

We can "ignore" the negative solution because a negative time doesn't make any physical sense.

Note that I chose to make my "y-down" the positive direction.

can u please explain why it doesnt work if I find the time it takes the ball to travel 14 m then make an equation in terms of t as follows: 4.9t^2= 7(t+sqrt(20/7)) — Philip O'Neill, Feb 10 '15 at 20:45

Objects falling.

2 Answers2