Will it make a difference?: Home: Capstone Project

Will it make a difference?

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So, a gravity constant of -9.7938 m/s2 doesn't seem to be that much of a difference. But, you never know until you check.

Say our good friend Derrick Rose, who has a 40 inch vertical jump, were to jump in this situation.

To find the height he can jump with a lower gravity, I must first find the velocity at which he jumps off of the floor.

If I convert inches to meters, I have everything I need to plug it into some kinematics equations. $40in\cdot\frac{2.54cm}{1in}=101.6cm\:or\:1.016m$

Plug it into the equation $v^2=v_{0^{^2}}+2a\left(x-x_0\right)$

At the top of his jump, v will equal 0. Thus, $0=v_{0^{^2}}+2\cdot-9.8\frac{m}{s^2}\left(1.016m\right)$ , and $v_0=4.462\frac{m}{s}$

Then, just plug it back into the same equation, but with the new gravity constant.

$0=\left(4.462\frac{m}{s}\right)^2+2\cdot-9.7938\frac{m}{s^2}\left(x\right)$ . The value of x is the new height he can jump. X turns out to be 1.017m

WOW!!!! IF ALL THE PLANETS ALIGN, DERRICK ROSE CAN JUMP A WHOLE 1mm HIGHER!!!!!

Well, that's a little anticlimactic. So, what would the gravity need to be if he wants to jump 60 inches? (1.5 times what he can right now)

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[{:content=>"So, a gravity constant of -9.7938 m/s2 doesn't seem to be that much of a difference. But, you never know until you check.\nSay our good friend Derrick Rose, who has a 40 inch vertical jump, were to jump in this situation.\n \nTo find the height he can jump with a lower gravity, I must first find the velocity at which he jumps off of the floor.\nIf I convert inches to meters, I have everything I need to plug it into some kinematics equations. <img class=\"equation_image\" title=\"40in\\cdot\\frac{2.54cm}{1in}=101.6cm\\:or\\:1.016m\" src=\"/equation_images/40in%255Ccdot%255Cfrac%257B2.54cm%257D%257B1in%257D%253D101.6cm%255C%253Aor%255C%253A1.016m\" alt=\"40in\\cdot\\frac{2.54cm}{1in}=101.6cm\\:or\\:1.016m\">\nPlug it into the equation <img class=\"equation_image\" title=\"v^2=v_{0^{^2}}+2a\\left(x-x_0\\right)\" src=\"/equation_images/v%255E2%253Dv_%257B0%255E%257B%255E2%257D%257D%2B2a%255Cleft%2528x-x_0%255Cright%2529\" alt=\"v^2=v_{0^{^2}}+2a\\left(x-x_0\\right)\">\nAt the top of his jump, v will equal 0. Thus,   <img class=\"equation_image\" title=\"0=v_{0^{^2}}+2\\cdot-9.8\\frac{m}{s^2}\\left(1.016m\\right)\" src=\"/equation_images/0%253Dv_%257B0%255E%257B%255E2%257D%257D%2B2%255Ccdot-9.8%255Cfrac%257Bm%257D%257Bs%255E2%257D%255Cleft%25281.016m%255Cright%2529\" alt=\"0=v_{0^{^2}}+2\\cdot-9.8\\frac{m}{s^2}\\left(1.016m\\right)\">, and   <img class=\"equation_image\" title=\"v_0=4.462\\frac{m}{s}\" src=\"/equation_images/v_0%253D4.462%255Cfrac%257Bm%257D%257Bs%257D\" alt=\"v_0=4.462\\frac{m}{s}\">\n \nThen, just plug it back into the same equation, but with the new gravity constant.\n<img class=\"equation_image\" title=\"0=\\left(4.462\\frac{m}{s}\\right)^2+2\\cdot-9.7938\\frac{m}{s^2}\\left(x\\right)\" src=\"/equation_images/0%253D%255Cleft%25284.462%255Cfrac%257Bm%257D%257Bs%257D%255Cright%2529%255E2%2B2%255Ccdot-9.7938%255Cfrac%257Bm%257D%257Bs%255E2%257D%255Cleft%2528x%255Cright%2529\" alt=\"0=\\left(4.462\\frac{m}{s}\\right)^2+2\\cdot-9.7938\\frac{m}{s^2}\\left(x\\right)\">. The value of x is the new height he can jump. X turns out to be <img class=\"equation_image\" title=\"1.017m\" src=\"/equation_images/1.017m\" alt=\"1.017m\">\nWOW!!!! IF ALL THE PLANETS ALIGN, DERRICK ROSE CAN JUMP A WHOLE 1mm HIGHER!!!!!\nWell, that's a little anticlimactic. So, what would the gravity need to be if he wants to jump 60 inches? (1.5 times what he can right now)", :section_type=>"rich_text"}, {:attachment_id=>24417717, :section_type=>"attachment"}]

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Capstone Project - Jumping and Gravity

Will it make a difference?

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