Why 't = 0' here in this equation in the case below?

[Indranil]

Homework Statement

As we know, If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then h = 0 + (1/2)gT2 (for vertical motion)
T = √2h/g
Why here 'ut' I mean t = 0, 'u (initial velocity should be zero but why "t = 0"? Could you explain, please

Homework Equations

The Attempt at a Solution

h = ut + (1/2)gT^2 where u = initial velocity, g = acceleration, t = initial time, T = finial time

 

[Doc Al]

Why here 'ut' I mean t = 0, 'u (initial velocity should be zero but why "t = 0"?

What makes you think t = 0?

 

[diredragon]

Homework Statement

As we know, If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then h = 0 + (1/2)gT2 (for vertical motion)
T = √2h/g
Why here 'ut' I mean t = 0, 'u (initial velocity should be zero but why "t = 0"? Could you explain, please

Homework Equations

The Attempt at a Solution

h = ut + (1/2)gT^2 where u = initial velocity, g = acceleration, t = initial time, T = finial time

The equation you wrote is not correct. The initial and final time part of the equation make no sense. In two experiments where the initial time is set differently with the same time duration of the experiment we get different results using your formula. You need to draw a coordinate system and look up equations of motion to see what is going on. You have not said it clearly but the initial velocity should have only the horizontal component (x-axis) and no vertical component (y-axis). If you look up the equation of motion you will see why the first part equals 0. Your first thought was that if $vt=0$ then $t=0$ since $u$ is non-zero from your point of view. That is not so as you will see by drawing the coordinate system and learning a little bit more about the projectile motion.

https://courses.lumenlearning.com/physics/chapter/3-4-projectile-motion/

 

[Doc Al]

h = ut + (1/2)gT^2 where u = initial velocity, g = acceleration, t = initial time, T = finial time

As diredragon points out, this is incorrect. It should be: $h = u t + (1/2) g t^2$, where t is the final time. Since the initial velocity is horizontal, u = 0, thus $h = (1/2) g t^2$.

 

[diredragon]

As diredragon points out, this is incorrect. It should be: $h = u t + (1/2) g t^2$, where t is the final time. Since the initial velocity is horizontal, u = 0, thus $h = (1/2) g t^2$.

In the original post $u$ was used as just the initial velocity, so i suggest using $u_y$ to point out the direction so as to not confuse the OP who doesn't seem to take into account the vector decomposition of the problem.

So, $u_x=u$ and $u_y=0$

 

[Doc Al]

Good points!

 

[Indranil]

Good points!

Now let me clear in the equation below which one is initial and which one is the final time? 't or T'
h = ut + (1/2)gT^2

 

[Indranil]

In the original post $u$ was used as just the initial velocity, so i suggest using $u_y$ to point out the direction so as to not confuse the OP who doesn't seem to take into account the vector decomposition of the problem.

So, $u_x=u$ and $u_y=0$

Could you explain please why 'Uy' = 0 here?

 

[haruspex]

Could you explain please why 'Uy' = 0 here?

because

body is projected horizontally

 

[Indranil]

because

Do you mean to say 'as the body projected horizontally, so, the vertical velocity is zero I mean Uy = 0'?

 

[haruspex]

Do you mean to say 'as the body projected horizontally, so, the vertical velocity is zero I mean Uy = 0'?

Yes, the initial vertical component of velocity is zero.

 

[diredragon]

Now let me clear in the equation below which one is initial and which one is the final time? 't or T'
h = ut + (1/2)gT^2

The object is moving with the initial velocity $u$ which has only a horizontal projection, in a $x-y$ plane under the influence of gravitational attraction on the y-axis.

The equation of motion which tells us the height of the object is: $y=u_yt - \frac{1}{2}gt^2$. I used that the acceleration is positive in the negative direction of the y-axis.

$y_f-y_i=u_y(t_f-t_i)-\frac{1}{2}g(t_f-t_i)^2$. If you draw the diagram you will see that:
$0 - h=-\frac{1}{2}g(T)^2$ Note that $t_f-t_i=T$ is the duration of the process.
$T=\sqrt{\frac{2h}{g}}$

 

[Doc Al]

Now let me clear in the equation below which one is initial and which one is the final time? 't or T'
h = ut + (1/2)gT^2

As has been explained several times, this equation is incorrect. The correct equation (see above post) only has the final time.

 

[Indranil]

The object is moving with the initial velocity $u$ which has only a horizontal projection, in a $x-y$ plane under the influence of gravitational attraction on the y-axis.

The equation of motion which tells us the height of the object is: $y=u_yt - \frac{1}{2}gt^2$. I used that the acceleration is positive in the negative direction of the y-axis.

$y_f-y_i=u_y(t_f-t_i)-\frac{1}{2}g(t_f-t_i)^2$. If you draw the diagram you will see that:
$0 - h=-\frac{1}{2}g(T)^2$ Note that $t_f-t_i=T$ is the duration of the process.
$T=\sqrt{\frac{2h}{g}}$

From the equation given above, still i have two question
1. Why 'yf = 0'?
and
2. why Uy (tf - ti ) = 0? Could you explain please?

 

[Doc Al]

From the equation given above, still i have two question
1. Why 'yf = 0'?

Measured from the ground, the final position is at y = 0.

and
2. why Uy (tf - ti ) = 0?

Because Uy = 0. (Initial velocity was horizontal.)

 

[CWatters]

Now let me clear in the equation below which one is initial and which one is the final time? 't or T'
h = ut + (1/2)gT^2

Neither.

The equation should be written like this...

h = ut + (1/2)gt²

Both t are the same t. The equation gives you the height at _any_ time = t.

So to find the height after say 3 seconds, replace _both_ t with 3...

h = 3u + (1/2)g3²

Assume u=0 and g = -10m/s/s

h = -(1/2)(10)3²
= -45m

On the other hand if you are given h = -45m and asked to find t then you have to solve a quadratic for which there are TWO solutions. Only one solution will normally make sense (for example the other might be negative).

To solve a quadratic rearrange the equation to the general form ...

ax² + bt + c = 0
or
(1/2)gt² + ut -h = 0

Then note that

a = (1/2)g = -5 m/s/s
b = u
c = -h = 45m

Then use the general solution for a quadratic.

t = {-b +/- (Sqrt(b² - 4ac)}/2a

Lets say we want to know how long it takes to fall 45m. The height at time t will be -45m. Putting values into the equation we get..

t = +/- Sqrt(4*5*45) / -10
= +/- 30/-10
= +/- 3

The correct solution is +3 seconds.