Projectile negative vertical displacement

[rudransh verma]

Homework Statement

A ball is launched from a catapult with an initial speed of 3.25 m/s at an 61.7 degrees above the horizontal. If the basket is 93 cm away from the catapult horizontally where should the basket be placed vertically so the ball lands in the basket?

Relevant Equations

$$s= ut+\frac 12 at^2$$

I know how it’s done. So let’s jump on the question. The displacement that came was negative 6 cm. I want to know how this formula describes perfectly negative displacements.
I don’t understand the mechanism. If I assume $ut$ to be distance traveled in absence of g then what is $-1/2gt^2$ doing to $ut$ ?

 

[haruspex]

Suppose you just let a stone drop. After time t, how far has it fallen? How would you write that as a vertical displacement, with up defined as positive?

 

[kuruman]

Watch this demonstration. The gun is initially aimed at the monkey.

 

[rudransh verma]

Suppose you just let a stone drop. After time t, how far has it fallen? How would you write that as a vertical displacement, with up defined as positive?

$$-s=ut-\frac12gt^2$$
Or
$$x-x0=ut-\frac12gt^2$$ where x<x0

 

[haruspex]

$$-s=ut-\frac12gt^2$$
Or
$$x-x0=ut-\frac12gt^2$$ where x<x0

I specified "drop", so u=0.
The second equation is right, but the first has the wrong sign. If up is positive the displacement is given by $s=-\frac 12gt^2$.
So do you understand what the $-\frac 12gt^2$ does? It takes the ut that operates without gravity and adds the effect of gravity.

But I may not have grasped what it is that bothers you.

 

[rudransh verma]

I specified "drop", so u=0.

O yes.

If up is positive the displacement is given by s=−12gt2.

but s should be -ve since it’s going downwards. How is my first eqn wrong? But if I take -ve sign then it will get canceled by $-\frac12gt^2$ so s is now +ve. I don’t think I get it.

It takes the ut that operates without gravity and adds the effect of gravity.

So can I say $-\frac12gt^2$ is a type of distance operator. Is $ut$ distance from origin and $-\frac12gt^2$ distance from where the v=0 to the final position of the body in free fall motion? When I take difference I get the distance between the initial and final position of body.

I generally get stuck on formulas on how they reveal the truth about the world around us.
This formula perfectly reveals the displacement as +,-,0 based on the time elapsed. How?

 

[haruspex]

but s should be -ve since it’s going downwards

Right, and $s=-\frac 12gt^2$ will make s negative. But you had $-s=-\frac 12gt^2$, which is the same as $s=\frac 12gt^2$, and will make s positive.

 

[rudransh verma]

Right, and $s=-\frac 12gt^2$ will make s negative. But you had $-s=-\frac 12gt^2$, which is the same as $s=\frac 12gt^2$, and will make s positive.

So when do we take -$s$. I guess when it’s given and we are asked to find something else. But we don't have to take $s$ as -$s$ when we have to calculate it. Right?

 

[haruspex]

So when do we take -s.

I do not understand your question. The formula is $s=ut+\frac 12 at^2$. It is not $-s=ut+\frac 12at^2$. The result for s may be positive or negative , but the formula remains the same.

 

[rudransh verma]

I do not understand your question. The formula is $s=ut+\frac 12 at^2$. It is not $-s=ut+\frac 12at^2$.

I mean when do we take -s. When we are given a value of s in a question and it says body falls down. When we need to find something else. Then we take s=-5 for example. But when we are to find s then we don’t assume it’s value as -s because that will produce wrong result.

 

[haruspex]

we don’t assume it’s value as -s

We don't write -s because it is not -s. It is s, whether the value of s is positive or negative.
You are confusing the sign placed on the variable s with the sign of a value of s.

 

[rudransh verma]

We don't write -s because it is not -s. It is s, whether the value of s is positive or negative.
You are confusing the sign placed on the variable s with the sign of a value of s.

Ok

 

[rudransh verma]

It takes the ut that operates without gravity and adds the effect of gravity.

So can I say $-\frac12gt^2$ is a type of distance operator. Is ut distance from origin and $−\frac12gt^2$ distance from where the $v=0$ to the final position of the body in free fall motion? When I take difference I get the distance between the initial and final position of body.

I generally get stuck on formulas on how they reveal the truth about the world around us.
This formula perfectly reveals the displacement as +,-,0 based on the time elapsed. How?