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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
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