anonymous
  • anonymous
A projectile is launched straight up from ground level. It reaches a maximum height of 42.9 m. Find the total time from it is launched until it lands on the ground again. (Ignore air resistance).
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
do you have the initial velocity?
anonymous
  • anonymous
because it would make more sense with the initial velocity but this is what I would have \[v ^{2}=v _{0}^{2}-2g \Delta y\] we knwo delat y is 42.9 g is about 10 and v is 0 since at its apex the velocity is 0
anonymous
  • anonymous
so \[0^{2}=v _{0}^{2}-2(10)(42.9)\]

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anonymous
  • anonymous
\[858=v _{0}^{2}\] \[v _{0}\approx 29.3\]
anonymous
  • anonymous
so now we have initial velocity since we know free fall motion is symetric and initial launch is equal to the velocity as it hits the ground
anonymous
  • anonymous
so now we can use the initial velocity to find time \[0=29.2-(10)t\] solve for t and we have the time to rise
anonymous
  • anonymous
multiply that by 2 and we have the total time since trise=tfall

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