Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
[tex]x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration[/tex]
So now we have an equation and unkown value.
for the thrown rock
[tex]\frac{1}{2}(9.8)*t^2+29*t-300=0[/tex]
for the dropped rock
[tex]\frac{1}{2}(9.8)*t^2+0*t-300=0[/tex]
solving both equation with the quadratic formula:
[tex]\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}[/tex]
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
