Think Fast

First of all, consider the point where the rock leaves the boy's hand as point $A$ and let the highest point where the velocity of the rock becomes equal to $0$ be point $B$ .

The time required for the rock to travel from $A$ to $B$ is always equal to the time required for the rock to travel from $B$ to $A$ . This is valid only when other resistive forces are neglected, i.e. when air resistance is neglected.

So, we calculate the time required for the rock to travel from $A$ to $B$ .

initial velocity = velocity at $A$ = $u$ = $30$ m/s

final velocity = velocity at $B$ = $v$ = $0$

acceleration = acceleration due to gravity = $a$ = $-9.8$ m/s/s

So, by Newton's first equation of motion,

$v = u + at$

$0 = 30 + (-9.8)t$

$t = \frac{30}{9.8}$

$t = 3.061$ sec

Hence, the total time for the rock to travel back to the same point is $2t = 2 \times 3.061 = 6.122$ sec

That's the answer!

Acceleration is equal to the change in velocity divided by the time taken:

$a (m/s^2) = \frac{v_{1} (m/s) - {v_2} (m/s)}{t (s)}$

If we imagine the rock's path, it makes sense to assume that's it's velocity will change from $30m/s$ to $0m/s$ as it travels upwards, and then $0m/s$ to $-30m/s$ on it's way back down. Hence the change in velocity is $30m/s$ to $-30m/s = -60m/s$ . Given this, and given that the acceleration of gravity is $-9.8m/s$ , we can plug some values into the acceleration formula:

$-9.8m/s^2 = \frac{-60m/s}{t (s)}$

Now this is simply a case of rearranging the equation to calculate $t$ :

$(-9.8m/s^2 = \frac{-60m/s}{t (s)}) \times t (s) =$

$(-9.8m/s^2 \times t (s) = -60m/s)$ divide by $-9.8m/s^2 =$ : $(t (s) = \frac{-60m/s}{-9.8m/s^2}) =$ $6.1224... = \frac{-60}{-9.8}$

The answer is $\boxed{6.12}$ !

$s=ut+\frac{1}{2}at^2$

$s=0, u=30, a=-g=-9.8$

$0=30t-\frac{1}{2}gt^2$

$0=t(30- \frac{1}{2}gt)$

$t=0$ or $(30- \frac{1}{2}gt)=0$

$30=\frac{1}{2}gt$

$t=\frac{60}{g}=\frac{60}{9.8}=\boxed{6.12} (3 s.f)$

First consider the time it takes for the rock to reach peak height. At the top of the trajectory the velocity of the stone rock is 0 m/s.

By using the equation: $t = \frac{v-u}{a}$

where:

• $v$ is the final velocity

• $u$ is the initial velocity

• $a$ is the acceleration

• $t$ is the time

Plugging in values of $v = 0$ , $u=30$ , $a=-9.8$ , we get $t = \frac{0-30}{-9.8} = 3.06s$

It takes the same amount of time to fall as it does to rise, so double the previous answer to get the final answer of $\boxed{6.12 s}$ to 3 s.f.

Many ways to answer the question, this was the first to occur to me.

t=2v/g

Vertical Up Motion

$v_{0}=30 m/s$

$v_{t}=0$

$g=9.8 m/s^{2}$

Then..

$v_{t}=v_{0} - gt$

$30=9.8t$

$30=9.8t$

$t=3.061$

You just can double it, so the answer is $\boxed{6.122}$

(time\quad of\quad flight)f=\frac { 2u }{ g } \ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 2*30 }{ 9.8 } \ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =6.122sec

We have to take time of flight which is 2*u/g. After substituting values we will get 6.122 sec

time of flight (t=2u/g)

time of flight of a body which is projected vertically upward with a velocity 'u' is 2u/g.so time here the height of rock and the persons height are neglected.so now 60/9=6.1

it is given that initial velocity(u)=30m/s at the highest point velocity of the stone become zero so , V=0 and g=9.8 m/s^2 So, by Newton's first equation of motion, v=u+at 0=30-9.8t t=3.061s 2t=6.122

easiest method is:- time of flight=2u/g, which gives 6.122 directly

Here, for upward journey v=u+at and t could be found. Double this t as time upw. = time downw. .

to find time we use first equation of motion v=u+at a= -9.8 u= 30 v= 0 0= 30 - 9.8t t=30/9.8 t= 3.06 as time would be same T=2t=6.12sec

For the upward journey:

v=u+at

0=30-9.81t

t= 30/9.81 = 3.06 s

Time for upward travel= Time for downward travel = 3.06 s

Total time = 3.06 + 3.06 = 6.12 s

total-space=0 (space to go up - space to go down); $s=v0 \times t-\frac{1}{2} \times g \times t^{2}$ Evidencing time: $0=t \times (v0-\frac{1}{2} \times g \times t)$ First solution: t=0 not acceptable; Second solution: $t=2 \times \frac{v0}{g}= 6,12 s$

for case 1: Vo= 30 m/s

V=0

V=Vo + gt ---> T1= 3.06 s---> (1)

Also: S=VoT + 1/2 gT^2 -----> S=45.918 m

For Case 2: Vo=0

S= VoT + 1/2gT^2 -----> T2 = 3.06 s

T = T1 + T2 = 6.12 s

vf=vi+at vi=30 vf=0 g=-9.8

I did these kind of questions frequently in physics last chapter. Since the rock moves up at 30m/s you just divide that by 9.8 to figure out when it hits 0 in terms of velocity. Then multiply that by two because gravity pulls the rock down at the same speed the rock was decreasing at which eventually becomes 30m/s after about 3 seconds. The first part of the process of throwing the rock was then flipped to become the 2nd process.

the answer is 6 s... 3 s when the rock reach the top and 3 s when the rock hit his head...

using 1st equation of motion, vf=vi+gt here vf= 0 and vi = 30 m/s and g = -9.8 m/s^2.... so actually the this t is from a single side so the total time will be 2t = 2(30/9.8) = 6.12 sec

time = $2 \times \frac{Vf - Vi}{g}$

$\frac{Vf - Vi}{g} = \frac{0 m/s - 30 m/s}{- 9.8 m/s^{2}}$

You get 3.06122449 seconds, then multiply to two.

$2 \times 3.06122449$ = 6.122 seconds.

900=20s s=45

45=30t-5txt t=6sec

Use Equation v=u+gt ; v=0 at highest point of trajectory so after solving we get t= 30/9.8 ; Now Double up this time to get the total time T=2*t = 6.12 sec

Step 1 :V=u+at Step 2: 0=30+(-9.81*t) Step 3: It takes double the time for the ball to go up and to get to your head.

He has all the time till the stone reaches the point of projection ,

-We can get the time taken by the stone to get back from, T=2u/g we get T=6.120s by substituting the given values.

It is given that initial velocity(u) is 30 m/s. We know that the acceleration due to gravity is -9.8 m/s^{2} The final velocity(v) will be 0. According to fist equation of motion, we have v = u+at 0 = 30+(-9.8t) {-30}/{-9.8})=t 3.06 = t

Now, the stone takes 3.06 seconds to reach to the top. It will take the same time to come down. So, 3.06+3.06=6.12 seconds

6.12

First of all, You have to discover the time that the rock gets the higher point. So : V = V0-g.t =>

V = the final vecolity; V0 = the inicial velocity; g = gravity; t = time.

So : 30 = 0-9,8.t => T = approximately 3,06 s.

We have to know that the time the rock need top get the higher point,its the same time she falls to the ground. so t+t = 6,120s

Hope help You.

Choose your coordinate axes such that the upward direction is the positive y axis and the origin is at the point of projection.

From the third equation of motion, we have,

$v^2-v_{0}^2=2gh$ . . . (i) where $v$ is the final velocity, $v_{0}$ is the initial velocity, $g$ is the acceleration due to gravity and $h$ is the height attained.

At the highest point of a vertically projected body, the velocity is zero. setting $v=0$ and substituting in (i), we get,

$h=\frac{-v_{0}^2}{2g}$ . . . (ii)

Applying the second equation of motion to the downward motion, we have

$-h = vt+\frac{1}{2}gt^2$ . . . (iii) where $t$ is time. The - sign is there in the left side as the displacement of the body is downwards.

For a vertically projected body, the time of ascent is equal to the time of descent. They are both equal to half of the time of flight. So, to obtain the time for which the rock stays in the air, it is easier to find the time of descent. When the body is descending the initial velocity becomes the final velocity of ascension i.e., $v=0$ . Substituting in (iii) and rearraginging, we get,

$t_{a}^2=-\frac{2h}{g}$ . . . (iv) where $t_{a}$ is the time of ascent. There's nothing wrong with the minus sign being on the right side as $g$ is negative. Substituting (ii) in (iv) we get,

$t_{a}=\sqrt{\frac{2v_{0}^2}{2g^2}}=-\frac{v_{0}}{g}$ . The - sign appears as g is negative and $\sqrt{x^2} = -x$ if $x<0$ .

Then, time of flight is given by,

$t_{f}=\frac{-2v_{0}}{g}$

Substituting $v_{0}=30 \mathrm{\, m \, s^{-1}}$ and $g = -9.8 \mathrm{\, m \, s^{-2}}$ , we get,

$t_{f}=\boxed{6.120} \mathrm{\, s}$

The problem can be solved by taking the axes in any other way; the answer obtained will be the same.

S = Vo.t + 1/2.g.t^2 ---> s = 0 + 1/2 .g.t^2 ---> vt = 0 + 1/2 .g.t^2 ---> vt = 1/2 .g.t^2 ---> v = 1/2.g.t ----> t = 2v/g ---> t = 2.30[m/s]/ 9,8 [m/s^2] = 60/9,8 s = 6,122448979.... = 6,12 s. Answer : 6.12E-0