# Velocity decreasing and time increasing

My question is in regards to forming linear relationships and solving through direct proportion. The question is as follows:

A ball is thrown vertically upwards with a velocity v of 12 m/s, which decreases by an amount proportional to t, the number of seconds where the ball moves upwards. After 0.5 seconds, the ball has slowed to a velocity of 7.1 m/s. How many seconds after it is thrown upwards will the ball start falling to the ground?

I tried solving through the basic v=kt l as in through direct proportion, but if I solve I get v=24t. And in that case, when v=0, that will only be when time is zero. So how do I get the formula right?

• The velocity is not proportional to $t$. The decrease is. – user296602 Feb 4 '18 at 3:59
• @user296602 i could sense that but I got lost. How do I form the relation? – TGamer Feb 4 '18 at 4:00

The velocity started at $12$ m/s, and after $0.5$ second, it was $7.1.$ The difference is $12 - 7.1 = 4.9,$ and this is the amount of the decrease in velocity in $0.5$ second.

In general, if we look at the ball after $t$ seconds, and it is moving at velocity $v,$ the decrease in velocity since it started moving upward is $12 - v.$

We are given that the amount of decrease in the vertical velocity of the ball is directly proportional to how long it has been moving upward. Since the decrease is directly proportional to $t$ there must be a constant $k$ such that $$12 - v = kt. \tag1$$

From the fact that the velocity was $v=7.1$ when the time was $t=0.5,$ by plugging those values into Equation $(1)$ we can solve for $k.$

Once we have $k,$ we can set $v = 0$ and solve for $t$ to find out the time when the ball has stopped moving upward.

Rather than speaking of the amount of decrease in velocity, I think it is more usual to speak of the change in velocity over a period of time. So if the velocity starts at $v = 12$ at time $t = 0,$ we say that $12$ is the initial velocity, and we call this $v_0.$ If the velocity at a later time is $v,$ the change is $v - v_0 = v - 12$; we always measure change by taking "final minus initial."

Since $v = 7.1$ when $t = 0.5,$ at that time the change in velocity has been $$v - v_0 = 7.1 - 12 = -4.9.$$ It's a negative number because the velocity is decreasing and because a decrease is represented by a negative change.

Since the decrease is proportional to time, so is the change (which is just exactly the opposite value). So we have $$v - 12 = Kt.$$

You can proceed to solve this just like when we had a formula for the decrease; it just happens that now the value of $K$ will be negative.

• (+1): This is a very nicely written answer that actually addresses the issue OP had without just using some kinematics formulae. – user296602 Feb 4 '18 at 16:02

Acceleration is $(7.1-12)/0.5=-9.8$, so it takes $12/9.8=1.22$ seconds to reach $0$ velocity.

$\star)$: $v_f=v_i -at;$

$f:$ final; $i$:initial;

$a$: acceleration(constant).

1) $v_f= 7.1[m/s]$; $v_i= 12[m/s]$; $t=0.5 [s].$

Solve for $a$ in $\star)$.

2) Set

$v_f=0$(The ball reverses direction);

$v_i =12[m/s].$

Solve for $t$ in $\star).$

For $t=0$ the ball goes upwards with initial velocity $v_0= 12\,\mathrm{ms^{-1}}$. So we have that $$v(t)=v_0-at\qquad \Longrightarrow\qquad a=\frac{v_0-v(t)}{t}$$ and after $t=t_1=0.5\,\mathrm s$ we have $v_1=v(t_1)=v(0.5)=7.1\mathrm{ms^{-1}}$ and then $a=9.8\mathrm{ms}^{-2}=g$ (i.e. $a=g$ is the acceleration due to gravity).

At the time $t=t^*$ when $v(t^*)=0$ the ball will stop and begin to fall down, that is $$v(t^*)=0=v_0-gt^*\qquad \Longrightarrow\qquad t^*=\frac{v_0}{g}\approx 1.22\mathrm s$$