How Do You Calculate the Takeoff Speed in a Long Jump Problem?

AI Thread Summary
To calculate the takeoff speed of a long jumper who leaves the ground at a 23-degree angle and travels 8.7 meters, one must resolve the initial velocity into horizontal and vertical components. The horizontal component can be expressed as v*cos(23), while the vertical component is v*sin(23). Using trigonometric relationships, the height of the jump can be determined from the tangent function, which allows for the calculation of the time of flight. With the time known, the initial velocity can be derived from the horizontal distance using the equation Xf = Vi*cos(23) * time. The correct takeoff speed is determined to be 11 m/s.
Lil'Physicist
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Homework Statement



A long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper?

Homework Equations



Trigonometric functions
?

The Attempt at a Solution



Recently started physics and am going over stuff from the last two tests. Got stuck on this one and needed some help.

None of the three standard kinematics equations allows me to solve this (I end up having two unknowns for any equation I choose), and trigging out the velocity in the beginning leaves me with a triangle that only has angle measures. How can I start on this problem?
 
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Denoting the takeoff velocity as v, resolve it into the horizontal and vertical components. You will be able to get an expression each for the horizontal and vertical components, which form a pair of simultaneous equations.
 
Hi Lil'Physicist. welcome to PF. It is a vey common problem in projectile motion. Open any reference book or visit Hyper Physics site and go through the projectile motion.
Try to find out two expression. (1) time of flight and (2) range of the projectile.
Then try to solve the problem.
 
Fightfish said:
Denoting the takeoff velocity as v, resolve it into the horizontal and vertical components. You will be able to get an expression each for the horizontal and vertical components, which form a pair of simultaneous equations.


I ended up with sin 23=v(y-axis)/v for the y-axis velocity and cos 23=v(x-axis)/v for the x-axis velocity. I do not understand which kinematics equation to substitute this into because of the two unknown variables.

If someone could show the first few steps of the problem, I need that to get the ball rolling. The correct answer is 11 m/s... now just to figure out how that is so...
 
Last edited:
Lil'Physicist said:
I ended up with sin 23=v(y-axis)/v for the y-axis velocity and cos 23=v(x-axis)/v for the x-axis velocity. I do not understand which kinematics equation to substitute this into because of the two unknown variables.

If someone could show the first few steps of the problem, I need that to get the ball rolling. The correct answer is 11 m/s... now just to figure out how that is so...

Using trig functions, you can find the maximum height the jumper will go...

tan23=(height of jump)/8.7

Now you have height of jump, you can find the time it takes to complete the jump. Once you have time, you have horizontal distance, time and then solve for Vi via equation:
Xf = Vicos(23) * (time) where Xf is the horizontal distance of the jump and Vi is the initial velocity of the jumper
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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