Projectile Motion long jumper Problems

[312213]

Homework Statement

An athlete executing a long jump leaves the ground at a 20° angle and travels 7.80 m.

(a) What was the takeoff speed?

Homework Equations

Trigonometry- 3.6tan(20)=1.42 which is the max height
Vy² = Vyo² + 2a(y-yo) to find initial velocity
Vy = Vyo + at to find time
x = xo + Vxot + 1/2at² to find velocity initial

The Attempt at a Solution

3.6tan(20)=1.4194839136381892092690867428297m is the meters for max height.

Vy² = Vyo² + 2(-9.8m/s²)(1.4194839136381892092690867428297)
Vyo² = 27.821884707308508501674100159461m/s
Vyo = 5.2746454579723657955166904897785m/s

Vy = 5.2746454579723657955166904897785m/s + (-9.8m/s)t
t = 0.53822912836452712199149902956923s

0.53822912836452712199149902956923s x 2 = 1.0764582567290542439829980591385s for total hang time

7.8m = 0m + Vxo1.0764582567290542439829980591385s + 1/2(0m/s²)(1.07646)²
Vxo = 7.2459846457039799434766682275793This is not the correct answer, or so I'm told. I also tried twice the velocity for 14.491969291407959886953336455159 but it is still incorrect.

I know it is something similar to this because there is another part, "(b) If this speed were increased by just 6.0 percent, how much longer would the jump be? "

For this the answer resulted to be 0.96407999999999999999999999998m, which is correct. I think this means that I am on the right track, but need to put the correct number choice. Any inputs on possible numbers would be appreciated.

edit:I was told of another number that could be possible, 10.9, but I don't know how this number was reached, or if it is correct. Just thought I'd put it out

 

[chronorec]

I'm not sure how you got this equation - 3.6tan(20)=1.42 but it's wrong
The max height is given by S_y=(u²sin²20)/2g, this is derived from the equation V_y² = V_y0² - 2gS_y and letting V_y be zero.
For maximum range you should use the equation, R=(u²sin(40)/g), this is derived using S=V₀t+0.5at² for both the x and y directions, then letting S_y be zero and eliminating t from the equations.

 

[312213]

I am not clear on what the "Sy=(u2sin220)/2g" equation is. I have never seen this before, nor do I understand what the S and the u and maybe the g represents.

Also I forgot clarify that the 3.6 is half of the total distance.
3.6tan(20) is supposed to be the meters upwards that would have been jumped, according to trigonometry. Thinking back, I think this part might be wrong, but it did, more or less, lead to a possible answer.

The 3.6 represents the distance the jumper reaches before he reaches his max height. Later the time is multiplied by 2 to account the fact that 3.6 is only half the total jump/distance/time.

edit: I tried with 7.8, and without multiplying the time by two. This got me, 10.247369758701774468445911215078 m/s. This might have something to do with the 10.9 I wrote in the edit of the opening post, and might of been rounded to result in 10.9.

 

[chronorec]

I'm using S to denote the displacement of the object and u to denote the initial velocity of the object, g is the acceleration due to gravity(ie. 9.81) and R is the maximum range which is 7.8 is this case.
Using 3.6tan(20) to denote the maximum height is wrong as you are assuming that the person is jumping in a triangle motion, but it's actually projectile motion which is a curved path.

 

[312213]

I see. I didn't understand your equations because I am being taught using a less proper way. Pardon my ignorance.

Aside from that, with the R=(u²sin(40)/g), I seems to have gotten the right answer.
Doing 7.8 = (u²sin(40)/9.8) gets me u = 10.905023123552279645808268515182m/s. This seems correct. I will be checking it later. Thank you.

 

[312213]

Since the 10.9 seems correct, I now run into a new problem. "If this speed were increased by just 6.0 percent, how much longer would the jump be?"

My previous method to attain the correct answer worked, but it worked under an incorrect velocity. It might have something to do with the way I played with the numbers but it won't work with the correct velocity.

What I would normally do, if I had not achieved the answer for the increased speed, but had the velocity:

10.9m/s×1.06 = 11.56m/s

With this I used trigonometry to find the time:

11.56tan(20) = 4.2m/s

(4.2m/s)/(9.8m/s²) = 0.43s

With this I would multiply by two to find the total hang time and then multiple with the velocity x.
0.43s×2 = 0.86s
0.86s×11.56m/s = 9.925m.

9.925m-7.8m does not equal the number that would be correct, which is 0.964m, or something similar.
Where does this go wrong, that is, if it is even right at all.The full format/numbers:

"If this speed were increased by just 6.0 percent, how much longer would the jump be?"

My previous method to attain the correct answer worked, but it worked under an incorrect velocity. It might have something to do with the way I played with the numbers but it won't work with the correct velocity.

What I would normally do, if I had not achieved the answer for the increased speed, but had the velocity:

10.905023123552279645808268515182m/s×1.06 = 11.559324510965416424556764626093m/s

With this I used trigonometry to find the time:

11.559324510965416424556764626093tan(20) = 4.2072500502151376623954096047118m/s
(4.2072500502151376623954096047118m/s)/(9.8m/s²) = 0.42931122961378955738728669435827s

With this I would multiply by two to find the total hang time and then multiple with the velocity x.
0.42931122961378955738728669435827s×2 = 0.85862245922757911477457338871653s
0.85862245922757911477457338871653s×11.559324510965416424556764626093m/s = 9.9250956386147591540094551406057m.

9.9250956386147591540094551406057m-7.8m does not equal the number that would be correct, which is 0.96407999999999999999999999998m, or something similar.
Where does this go wrong, that is, if it is even right at all.

 

[chronorec]

Try using back the equation R=(u²sin(40)/g), notice that the maximum distance is directly proportion to u², hence when the velocity is changed to 1.06 times, R will become 1.06² times.

 

[312213]

Ah thanks. I see know. Okay then I guess that wraps up everything. Only thing left is to fully understand those equations and fit it to the ones I learned. That is all.