Question

A marble is projected vertically up by a spring gun, and reaches the maximum height of 9.8m. What is the initial speed of the marble? How long did it take the marble to reach maximum height?
So far I've got x= 9.8m
Vi=?
g=-9.8 m/s
I can't get any further :/

Answer 1

Well starting off with \[a=-g\]
Integrating that we get.
\[v=-(g*t)+v _{0}\] Where \[v _{0}\] is the inital velocity of the marble.
Integrating again we get the position as a function of time being:
\[x=-\frac{ g*t ^{2} }{2 } + v _{0}t +x _{0}\]
Where x_0 is the initial position of the marble so we'll let it be zero.
To get the maxium height that the marble reaches we differentiate x and set it equal to zero; ie, let velocity be zero.

\[0=-9.8*t _{m} +v _{0}\] Where t_m is the time when the marble reaches its maximum height.
\[g*t _{m}=v _{0}\]
Since that at t=t_m x=x_m = 9.8, subbing into the position equation
\[x _{m}=\frac{ -g*t _{m} ^{2} }{ 2 } + g*t _{m}^{2}\] Simplifing
\[x _{m}=\frac{ g*t _{m} ^{2} }{ 2 } \]
Rearanging for t_m
\[t _{m}=\sqrt{\frac{ 2x _{m} }{ g }}\]
So then just subbing in g and the maximum height (x=9.8m) you can get the time it takes to reach the maximum height and using that you will sub it into
\[g*t _{m}=v _{0}\] and then you have your initial velocity of the marble.
All the best :)