What Speed Does the Athlete Land in Long Jump?

AI Thread Summary
The discussion focuses on calculating the landing speed of an athlete in a long jump, given a launch angle of 25 degrees and a distance of 8.8 meters. Participants reference the range formula R = (v0^2 sin(2θ)) / g to find the initial velocity (v0). To solve for v0, the formula is rearranged to v0 = √(xg/sin(2θ)). There is confusion among participants regarding the substitution and manipulation of the formula, particularly in obtaining the correct initial velocity. The conversation emphasizes the importance of correctly applying the range formula to determine the athlete's landing speed.
zman2011
Messages
7
Reaction score
0

Homework Statement


During a high school track meet, an athlete performing the long jump runs and leaps at an angle of 25 degrees and lands in a sand pit 8.8 m from his launching point. If the launch point and landing point are at the same height, y=0m, with what apeed does the athlete land?



Homework Equations





The Attempt at a Solution


Im completely stumped. any help would be appreciated.
 
Physics news on Phys.org
Are you familiar with the range formula?
 
no i am not.
 
can anyone tell me what the range formula is?
 
R = ((v0)2 sin(2\theta0)) / g

Rearranging this formula and solving for v0 will give you the initial velocity. You can figure out the final velocity from there.
 
well then how do i find initial velocity to put it in.
 
i didnt see the second part of that. sorry. thank you
 
im having trouble with this. i keep getting initial velocity as a decimal
 
Maybe you aren't substituting/manipulating properly.

x = \frac{v_o^2sin2\theta}{g}
v_o=\sqrt{\frac{xg}{sin2\theta}}
 
Back
Top