# Calculating Velocity when Stuntman Jumps from 1.25m Height

• Istiak
In summary: And this gives me no confidence that you understand the concept of a gravitational...I don't understand what you are asking. Do you know what vector components are?Do you know what vector components are?
Istiak
Homework Statement
A stuntman jumped from $1.25 \ \text{m}$ height and, landed at distance $10 \ \text{m}$. Find velocity when he jumped. (Take $\text{g}=10 \ ms^{-2}$)
Relevant Equations
$$h=\frac{1}{2}gt^2$$
>

>A stuntman jumped from $1.25 \ \text{m}$ height and, landed at distance $10 \ \text{m}$. Find velocity when he jumped. (Take $\text{g}=10 \ ms^{-2}$)

I had solved it following way.

$$h=\frac{1}{2}gt^2$$
$$=>1.25=5\cdot t^2$$
$$=>t=\frac{1}{2}$$
And, $$s=vt$$
$$v=\frac{s}{t}$$
$$=\frac{10 \ m}{\frac{1}{2} \ s}$$
$$=20 \ m/s$$

The answer is correct (checked from book answer). But, $v$ is average speed in the following equation.

$$s=vt$$

But, they told me to find initial velocity. That's why I think my answer is correct but, method is wrong so, the whole work is wrong either.

There's an assumption here that he jumped horizontally.

You've used, perhaps without realising, that the horizontal component of the velocity is constant.

Use those to get the right answer for the right reason!

PS you need to include units in your working. E.g. ##20m/s##

PeroK said:
You've used, perhaps without realising, that the horizontal component of the velocity is constant.
That should be 9.8 m/s but, which formula I should use?

Istiakshovon said:
That should be 9.8 m/s but, which formula I should use?
Where did ##9.8m/s## come from? Formula for what?

PeroK said:
Where did ##9.8m/s## come from? Formula for what?
9.8 from Gravitational acceleration. Formula for solving that problem

Istiakshovon said:
9.8 from Gravitational acceleration. Formula for solving that problem
I don't understand what you are asking. Do you know what vector components are?

PeroK said:
Do you know what vector components are?
Yes! Ohh! I have to use Vector.

@PeroK I was trying to find distance of hypotenuse.
$$c=\sqrt{a^2+b^2}$$
$$=\sqrt{10^2 \ m+1.25^2 \ m}$$
$$=10.0778 \ m$$
I was thinking that initial velocity is $$0 \ ms^{-1}$$ since, that stuntman started his bike in initial position. Hence, initial velocity is $$0$$. But, OP says that I have to find initial velocity. How can I find? I was claiming that final velocity is $$0$$. But, If I use $$v^2=u^2+2gh$$ than, I get initial velocity is negative which isn't. So, what should I do with distance of that stuntman was "flying"?

We need to see some vector components of velocity.

Your method is correct and your answer is correct, however it seems to me that you haven't realized some things:
1. when you write ##s=vt##, s is the horizontal distance and v is the horizontal component of the initial velocity (and it happens to be that the initial velocity has only horizontal component but it could have been different), which does not change through the motion, that's why we can write ##s=vt## after all, this equation is when the velocity is constant (or when the velocity is the average velocity but this is not the case here).
2. The trajectory of the cyclist will be parabolical. In any case it won't be the hypotenuse (dotted line).
3. The total velocity of the cyclist during the parabolical trajectory changes both in magnitude and direction, however the horizontal component of the total velocity remains constant. It is the vertical component that keeps changing.

PeroK and Istiak
@Istiakshovon let me please ask you something: Why do you think the horizontal component of velocity remains constant through out the motion?

Delta2 said:
@Istiakshovon let me please ask you something: Why do you think the horizontal component of velocity remains constant through out the motion?
Cause, gravitational force is pulling.

Istiakshovon said:
Cause, gravitational force is pulling.
you need to be more precise than that. Ok gravitational force is pulling so what?

PeroK
Delta2 said:
you need to be more precise than that. Ok gravitational force is pulling so what?
gravitational force is pulling and, gravitational force is constant. Hence, horizontal component of velocity remains constant (To my mind, it's not constant completely cause, $v$ is increasing per sec).

Isn't that correct?

Nope its not because gravitational force is constant.

Delta2 said:
Nope its not because gravitational force is constant.
So, that's constant how?

A velocity component remains constant if the net force in the direction of that component is zero (that is a corollary from Newton's 2nd law). So do we have a net force in the horizontal direction?

Delta2 said:
do we have a net force in the horizontal direction?
If you are asking at t=0 s than, I will say, "yes" (I meant no. But, we have which is 0 N). Cause, at t=0 s only gravitational force is pulling that cyclist. hence, net force at t=0 is 0.

Istiakshovon said:
So, that's constant how?
Reading your posts on this thread gives me no confidence that you understand what a vector is. Your knowledge of the laws of motion seems limited to one dimensional motion.

You need to prove me wrong!

Delta2
Istiakshovon said:
If you are asking at t=0 s than, I will say, "yes" (I meant no. But, we have which is 0 N). Cause, at t=0 s only gravitational force is pulling that cyclist. hence, net force at t=0 is 0.
And this gives me no confidence that you understand the concept of a gravitational force!

Istiakshovon said:
If you are asking at t=0 s than, I will say, "yes" (I meant no. But, we have which is 0 N). Cause, at t=0 s only gravitational force is pulling that cyclist. hence, net force at t=0 is 0.
ok so we can agree that the net horizontal force at t=0 is 0. What about t=0.1 is there any net horizontal force?What about any t? is there any net horizontal force?

PeroK said:
You need to prove me wrong!
How can I? You know more than me :D .

PeroK said:
Your knowledge of the laws of motion seems limited to one dimensional motion.
Maybe, yes maybe, no. I am saying no, cause I had lot of problems of two dimensional motion earlier (that was lot of time ago nearly 7-8 months).

Delta2 said:
is there any net horizontal force?
Yes

Istiakshovon said:
Yes
I disagree with this but why do you say yes , which is this horizontal force you see?

Delta2 said:
I disagree with this but why do you say yes , which is this horizontal force you see?
sh*t! When I saw the message then, I was thinking that I am confusing with horizontal and, vertical. After using Google Translate, I am now sure that I was confused of horizontal and vertical axis. Now, I understand. Usually, I was talking about Vertical axis not Horizontal. :( . And, Sorry to disturb. Thanks to clear my confusion. :)

Delta2
I was mistaking another thing which is vertical force exists at t=0 s either. That's not 0 at t=0 s.

Istiakshovon said:
I was mistaking another thing which is vertical force exists at t=0 s either. That's not 0 at t=0 s.
yes the vertical force is not zero at all times, not only t=0. It is equal to the gravitational force.

Istiak
And, horizontal force is 0 N. But, If I think of hypotenuse then, there is force (maybe there is no force where hypotenuse ends).

Istiakshovon said:
But, If I think of hypotenuse then, there is force
Yes there is force along the hypotenuse, there is a component of the gravitational force along that way. But we usually talk about this component when we have an inclined plane, here we have a different case, and as I told you before the trajectory of the cyclist is not along the hypotenuse.

PeroK and Istiak

## 1. How is velocity calculated when a stuntman jumps from a height of 1.25m?

Velocity is calculated by dividing the change in height by the change in time. In this case, the change in height is 1.25m and the change in time is the time it takes for the stuntman to reach the ground.

## 2. What is the formula for calculating velocity in this scenario?

The formula for calculating velocity is: velocity = change in height / change in time. In this case, velocity = 1.25m / time.

## 3. How can the time it takes for the stuntman to reach the ground be measured?

The time can be measured using a stopwatch or a high-speed camera. The stopwatch should be started as soon as the stuntman leaves the ground and stopped as soon as they reach the ground. The high-speed camera can be used to record the jump and the time can be calculated by analyzing the footage.

## 4. Is there a specific unit for velocity?

Yes, velocity is typically measured in meters per second (m/s) or kilometers per hour (km/h).

## 5. Can other factors affect the velocity of the stuntman's jump?

Yes, there are several factors that can affect the velocity of the stuntman's jump, such as air resistance, the angle of the jump, and the weight and size of the stuntman. These factors should be taken into consideration when calculating the velocity.

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