# Calculate initial speed of an athlete - projectile motion

• Peter Halsall
In summary: T)=3m ... y(0)=0 what is the equation for y at time 0?In summary, at time 0 the jumper is at the ground. At time T she has jumped 3.0 m and is at x=3.0.
Peter Halsall

## Homework Statement

An athlete can cover 3.0 m for a standing long jump. Assuming that this athlete jumps at an angle of 45 degrees to the horizontal. calculate the speed at which the jumper can launch herself from a standing start.
Assume g =10 ms^-2
and no air resistance
Assume that horizontal velocity never changes

## Homework Equations

v=u+a*t
x=(u+v/2)*t
x=u*t+(a/2)*t^2
x=v*t-(a/2)*t^2
v^2=u^2+2*a*x
v= final velocity
x= displacement
a= acceleration
u= initial velocity
t= time
sin theta*initial velocity= vertical initial velocity
cos theta*initial velocity= horizontal initial velocity

## The Attempt at a Solution

In my scientific calculator I observed that cos 45 = sin 45 and thus the initial vertical and horizontal velocity must be equal to each other. From my observations, I do not think I have enough information to find this from the formula's listed. I don't really know what to do from here =/

Peter Halsall said:

## Homework Statement

An athlete can cover 3.0 m for a standing long jump. Assuming that this athlete jumps at an angle of 45 degrees to the horizontal. calculate the speed at which the jumper can launch herself from a standing start.
Assume g =10 ms^-2
and no air resistance
Assume that horizontal velocity never changes

## Homework Equations

v=u+a*t
x=(u+v/2)*t
x=u*t+(a/2)*t^2
x=v*t-(a/2)*t^2
v^2=u^2+2*a*x
v= final velocity
x= displacement
a= acceleration
u= initial velocity
t= time
sin theta*initial velocity= vertical initial velocity
cos theta*initial velocity= horizontal initial velocity

## The Attempt at a Solution

In my scientific calculator I observed that cos 45 = sin 45 and thus the initial vertical and horizontal velocity must be equal to each other. From my observations, I do not think I have enough information to find this from the formula's listed. I don't really know what to do from here =/
There are two dimensions to be considered. You can apply those SUVAT equations separately in each.
In general, a SUVAT equation involves four variables of initial speed, final speed, acceleration, distance covered, and time taken. So given any three you can find the other two, i.e. there are effectively six unknowns altogether forthe whole motion.
In this case, you know the distance and acceleration in each direction. That settles four of them. You also know the initial velocities are the same in each direction, so that's one more equation. What else is the same for both motions?

Peter Halsall said:
...In my scientific calculator I observed that cos 45 = sin 45 and thus the initial vertical and horizontal velocity must be equal to each other ...
So can you produce an equation for the horizontal component and another for the vertical component, then equate them to find another piece of data?

haruspex said:
There are two dimensions to be considered. You can apply those SUVAT equations separately in each.
In general, a SUVAT equation involves four variables of initial speed, final speed, acceleration, distance covered, and time taken. So given any three you can find the other two, i.e. there are effectively six unknowns altogether forthe whole motion.
In this case, you know the distance and acceleration in each direction. That settles four of them. You also know the initial velocities are the same in each direction, so that's one more equation. What else is the same for both motions?
How do we know the acceleration for the horizontal component? And what do you mean by 'initial velocities are the same in each direction, so that's one more equation' ?

Peter Halsall said:
How do we know the acceleration for the horizontal component? And what do you mean by 'initial velocities are the same in each direction, so that's one more equation' ?
...Assume that horizontal velocity never changes
If horizontal velocity never changes, what is horizontal acceleration?

UH = UV is a simple, but valid equation.

note: sin(45deg) = cos(45deg) = 1/√2
so if the initial speed is u, the vertical and horizontal components are u/√2

How do we know the acceleration for the horizontal component?
Newton's Laws ... what are the forces acting on the jumper?
Alternately - if you are assuming the horizontal velocity does not change, then you can just use the definition of acceleration.

And what do you mean by 'initial velocities are the same in each direction, so that's one more equation' ?
He means the components of the initial velocity:
The initial velocity would be ##\vec v(0)=v_x(0)\hat\imath + v_y(0)\hat\jmath## but it is easier to work in components.
At time t>0, ##\vec v(t)=v_x(t)\hat\imath + v_y(t)\hat\jmath## ... this just says you can always break the velocity into horizontal and vertical components.
##v_x(0)=v_y(0)## is an equation that says "the components of the initial velocity are the same as each other" - maths is a language.

You are going to end up solving simultanious equations - so just write down all the relationships you know that apply to the situation.
i.e. if it takes time T to make the jump, then x(0)=0 and x(T)=3m ... y(0)=0 what is y(T)?

How do I know the displacement vertically? and I don't know what else is the same in terms of both motions. They both have the same initial velocity but the difference in acceleration (which for horizontal it must be 0, thank you) implies that the motions must be different.

Peter Halsall said:
How do I know the displacement vertically? and I don't know what else is the same in terms of both motions. They both have the same initial velocity but the difference in acceleration (which for horizontal it must be 0, thank you) implies that the motions must be different.
Is the landing point higher than the take-off point? Lower?
Think through the five variables I listed for each of the two directions, horizontal and vertical. Which ones are the same in both directions?

haruspex said:
Is the landing point higher than the take-off point? Lower?
Think through the five variables I listed for each of the two directions, horizontal and vertical. Which ones are the same in both directions?
Time? Time can't be different for either directions of motion, the direction has no effect on it whatsoever.

Peter Halsall said:
Time? Time can't be different for either directions of motion, the direction has no effect on it whatsoever.
That's right, the two motions occur over the same period of time. So what equations can you now write?

haruspex said:
That's right, the two motions occur over the same period of time. So what equations can you now write?
x=u*t+(a/2)*t^2

Peter Halsall said:
x=u*t+(a/2)*t^2
Is that for the horizontal or the vertical direction? You need one equation for each, using distinct symbols where appropriate.

haruspex said:
Is that for the horizontal or the vertical direction? You need one equation for each, using distinct symbols where appropriate.
Could you use it for both?

Peter Halsall said:
Could you use it for both?
You can use the same generic equation, but the variables may take different values, so use a distinct name for each variable that might have a different value.

haruspex said:
You can use the same generic equation, but the variables may take different values, so use a distinct name for each variable that might have a different value.
Ok so I have two equations
Vertical component:
0=1/√2*u*t-5*t^2
Horizontal component:
3=1/√2u*t

From this I can rearrange the horizontal equation to get t by itself and then substitute it into the vertical equation to attain u?

Peter Halsall said:
Ok so I have two equations
Vertical component:
0=1/√2*u*t-5*t^2
Horizontal component:
3=1/√2u*t

From this I can rearrange the horizontal equation to get t by itself and then substitute it into the vertical equation to attain u?
Yes.

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air that is subjected only to the force of gravity. This type of motion can be seen in activities such as throwing a ball or shooting a projectile.

## 2. How do you calculate the initial speed of an athlete in projectile motion?

The initial speed of an athlete in projectile motion can be calculated using the equation: v0 = (d - (1/2)g(tf)2) / tf, where v0 is the initial speed, d is the horizontal distance traveled, g is the acceleration due to gravity, and tf is the time of flight.

## 3. What is the acceleration due to gravity?

The acceleration due to gravity is a constant value that represents the rate at which objects fall towards the Earth. On Earth, this value is typically given as 9.8 meters per second squared (m/s2).

## 4. Can you calculate the initial speed of an athlete without knowing the time of flight?

No, the time of flight is a necessary component in calculating the initial speed of an athlete in projectile motion. Without this value, the equation to calculate the initial speed cannot be solved.

## 5. How can the initial speed of an athlete affect their trajectory in projectile motion?

The initial speed of an athlete directly affects their trajectory in projectile motion. A higher initial speed will result in a longer horizontal distance traveled and a higher peak height, while a lower initial speed will result in a shorter horizontal distance traveled and a lower peak height.

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