Question

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).

Answer 1

do you have the initial velocity?

Answer 2

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

Answer 3

so
\[0^{2}=v _{0}^{2}-2(10)(42.9)\]

Answer 4

\[858=v _{0}^{2}\]
\[v _{0}\approx 29.3\]

Answer 5

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

Answer 6

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

Answer 7

multiply that by 2 and we have the total time since
trise=tfall