Solving a Long Jump Problem: Find Takeoff Speed & Increase Jump Distance

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Homework Help Overview

The discussion revolves around a long jump problem involving projectile motion, specifically focusing on calculating the takeoff speed of an athlete and the impact of increasing that speed on jump distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss breaking the problem into horizontal and vertical motion components. Questions arise regarding the use of the given angle and horizontal displacement to derive necessary variables. There is also an exploration of time of flight and calculations involving vertical velocity.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to approach the problem. Some calculations have been attempted, but there are questions about the validity of certain assumptions and methods used in those calculations.

Contextual Notes

Participants note the challenge of working with limited information, specifically the reliance on the horizontal displacement and the angle of takeoff. There is an acknowledgment of the need to clarify assumptions made during calculations.

jonnyboy261
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hey guys imm new here
i got a question in my book i can't figure out
it says:
an athlete executing a long jump leaves the ground at a 30 degree angle and travels
7.80m (a) what was the takeoff speed? (b) if this speed were increased just 5.0 % how much longer would the jump be?

anyone kno how to do this
i tried using the kinematic equations but ther isn't enuf info
please help if u know the answer.
thanks a lot !
 
Last edited:
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Start out by splitting the problem up into 2 parts: Horizontal motion & Vertical motion. From here, you will have 2 equations with 2 unknowns.
 
i know to do that.. but i still don't get how u can use the equations when u only have the horizontal displacement of 7.80 m ? how do i use the 30 degree angle to find anything?
 
You can calculate the time of flight.

If you have a horizontal displacement of 7.8 metres you can use symmetry to find where the highest point is - this is also half way through the flight.

The vertical velocity here is zero.

The 30 degree angle let's you split the flight into its horizontal and vertical components.
 
ok so i plugged 3.9 m and 0m/s into the displacement equation so it makes t=.892s
and then i found out Vi for vertical is 8.74 m/s
then i used SOH to find the hyponuse to be equal to 17.48 by dividing 8.74 by sin30
is this the right answer?
 
Last edited:
Why did you use 0m/s?
 

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