Projectile motion long jumper help

In summary, the conversation discusses the problem of finding the time and maximum height of an Olympic long jumper who jumps 7.7m with a horizontal speed of 8.8m/s. The conversation covers various equations and methods for solving the problem, including the use of projectile motion equations and finding the angle of the jump. The final solution involves using the equation Y = xtan(theta) - (g)x^2/2(V0^2cos^2(θ) to find the angle and solve for the maximum height. The conversation ends with a reminder to never give up hope when solving difficult problems.
  • #1
Amel
43
0

Homework Statement



An Olympic long jumper is capable of jumping 7.7m. Assuming his horizontal speed is 8.8m/s as he leaves the ground, how long is he in the air? Assume that he lands standing upright--that is, the same way he left the ground.



Homework Equations



Y = xtan(theta)0 - ((g)/2V02cos2(theta)0)X2

Vx0 = V0cos(theta)t

And other that I attempted at using...


The Attempt at a Solution



Ok so I have tried several things but, I don't have the angle at which he jumps to figure out the time, and you only get the initial velocity for the horizontal direction. I was going to use

Vx0 = V0cos(theta)t

to solve for Cos(theta) and figure out my angle but I don't even Have V, only Vx so am I suposed to assume they are the same?

If someone can point me in the right direction, this is the only problem I couldn't figure out earlier today.
 
Physics news on Phys.org
  • #2
Also I can't remeber exactly all the work I did today my work is kind of unorganized and its all over the place but I came to the conclusion that t was 1.11 but that was wrong.
 
  • #3
Vx0 = V0cos(theta)t

are you sure of this?, I don't think there's a "t" in that equation
 
  • #4
Oh hold up I mixed up two equation when writing this they were next to each other on my paper.

I got Vx0=V0cos(theta)

and X = V0cos(theta)t

my bad.
 
  • #5
The only acceleration that the long jumper is subject to is the acceleration due to gravity then no acceleration acts to oppose the long jumpers horizontal motion. Therefore his horizontal velocity remains constant. This problem simply breaks down to d/v=t. 7.7m/(8.8m/s)=.875s
 
  • #6
Wow so simple, I didn't even think about it like that I was trying to do it like the other projectile motion problems I did today. Thank you.
 
  • #7
Alright quick question... Is V the same as Vx0 because there is acceleration in the y direction so there is a vetical velocity component right? Because I am trying to now figure out how hight he goes.

So I figure I use Y= V0sin(theta)t- (1/2)gt2 but I don't have an angle. Which I thoguth I could figure out with X = V0cos(theta)t but I don't know V I only know Vx.
 
  • #8
no! that'll be true only if his motion was a straight one [as running]. does 7.7 m stands for the height y or the horizontal distance x? if the later, it'll be direct. use the last two equations you’ve stated in your later post to find t, that is:


Vx0=V0cos(theta)

and

X = [V0cos(theta)]t
 
  • #9
7.7m is the distance he jumps in the X direction.
 
  • #10
great, what's t value?
 
  • #11
The time the way he gave me was correct (we input our answers online, and it accepted it as correct) So I know the time, but how do I get the height without having an angle he jumps at, and I don't know how to get the angle without V since I only have Vx
 
  • #12
you have x,Vo and t, use the 2nd eq. to get the angle
 
  • #13
The jumper will reach his maximum height at t/2. For a hint. If you need more help let me know.
 
  • #14
You only have V[0][x] not V[0]
 
  • #15
So in this problem is V = Vx? If that's so then

X = [V0cos(theta)]t

Then the I would divide by t and V0 right? which gives me .9943 = cos(theta) so then inverse cos of .9943 would give me my angle? which I got to be 6.11 which doesent seem right, but I then used that and the Y I got was incorrect.

ok just saw the post that says you got Vx not V and then how do I find the angle without V?
 
  • #16
I have two attempts left on the problem and I don't know how to find the angle. So I can't guess anymore, need some hints. I mean I know if you want to maximize your X you would jump at a angle of 45 but not sure if you can assume that here and don't want to waste my attempt trying it.
 
  • #17
Ok on my last attempt, someone please help out before I screw it up. How can you find the angle if you don't know V0?
 
  • #18
Damn used up my last attempt and wasn't able to get the maximum height.
 
  • #19
type the eq. of Vy!
 
  • #20
The projectile equation is
Y = xtan(θ) - (g)x^2/2(V0^2cos^2(θ)
Τhe net displacement is zero. Hence
xtan(θ) = (g)x^2/2(V0^2cos^2(θ) or
tan(θ) - (g)x/2(V0^2cos^2(θ) or
tan(θ) = gx/2(vx)^2
Substitute the values and find θ.
 
Last edited:
  • #21
OR, using the 2nd eq. →Vo=Vxo/cos(theta)
you can substitute this into the Vy eq. which is
Vy= Vo sin(theta) – gt
you'll get
Vy= Vxo tan(theta) – gt
now you can calculate the value of theta, by lodging the value of (t/2) where at the maximum height the value of Vy is zero,


you should never give up hope Amel!:tongue:

Amel means hope, right?
 
Last edited:

What is projectile motion and how does it relate to long jumping?

Projectile motion is the motion of an object that is projected into the air and then falls under the influence of gravity. In long jumping, the athlete's body is the projectile and the motion is affected by gravity, air resistance, and other external factors.

What factors affect the distance of a long jump?

The distance of a long jump is affected by factors such as the angle and velocity at which the athlete takes off, the height of the jump, the strength and technique of the athlete, and external factors such as wind speed and direction.

How can I calculate the optimal angle and velocity for a long jump?

The optimal angle and velocity for a long jump can be calculated using the principles of projectile motion and kinematics equations. This calculation requires measurements of the athlete's physical abilities and data from previous jumps.

How does air resistance affect a long jump?

Air resistance, also known as drag, can have a significant impact on the distance of a long jump. As the athlete moves through the air, they experience resistance from the surrounding air which can slow them down and decrease the distance of their jump.

What are some techniques for improving long jump distance?

Some techniques for improving long jump distance include proper training and conditioning, focusing on technique and form, optimizing takeoff angle and velocity, and adjusting for external factors such as wind and weather conditions. Analyzing and learning from past jumps can also help improve performance.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
530
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
2K
Replies
3
Views
3K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
33
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
2
Replies
53
Views
3K
Back
Top