Answer:
The value of
such that the two stones reaches the same maximum height is
![v_{0}=\frac{\sqrt{913} }{2} \frac{ft}{s}](https://tex.z-dn.net/?f=v_%7B0%7D%3D%5Cfrac%7B%5Csqrt%7B913%7D%20%7D%7B2%7D%20%5Cfrac%7Bft%7D%7Bs%7D)
Explanation:
For the first stone, we need to find the maximum height reached, and for that we have to derivate the given position function
![f(t)=-16t^{2}+30t+43](https://tex.z-dn.net/?f=f%28t%29%3D-16t%5E%7B2%7D%2B30t%2B43)
<em>derivating</em>, we get
![f'(t)=-32t+30](https://tex.z-dn.net/?f=f%27%28t%29%3D-32t%2B30)
now we have to equalize the derivate to zero, and clear t
![0=-32t+30](https://tex.z-dn.net/?f=0%3D-32t%2B30)
![t=\frac{30}{32}s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B30%7D%7B32%7Ds)
then, if we put this value of t in the position function, we obtain that the maximum height for the stone is
![f_{max}=\frac{913}{16}ft](https://tex.z-dn.net/?f=f_%7Bmax%7D%3D%5Cfrac%7B913%7D%7B16%7Dft)
For the other stone, we have the given position function
![f(t)=-16t^{2}+v_{0}t](https://tex.z-dn.net/?f=f%28t%29%3D-16t%5E%7B2%7D%2Bv_%7B0%7Dt)
And again, <em>derivating</em> and clearing t, we obtain
![t=\frac{v_{0}}{32}s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv_%7B0%7D%7D%7B32%7Ds)
As the height must be the same for both stones, we can substitute in the position function theese values
![\frac{913}{16}=-16(\frac{v_{0}}{32})^{2}+v_{0}(\frac{v_{0}}{32})](https://tex.z-dn.net/?f=%5Cfrac%7B913%7D%7B16%7D%3D-16%28%5Cfrac%7Bv_%7B0%7D%7D%7B32%7D%29%5E%7B2%7D%2Bv_%7B0%7D%28%5Cfrac%7Bv_%7B0%7D%7D%7B32%7D%29)
From where our value for
results to be
![v_{0}=\frac{\sqrt{913} }{2}\frac{ft}{s}](https://tex.z-dn.net/?f=v_%7B0%7D%3D%5Cfrac%7B%5Csqrt%7B913%7D%20%7D%7B2%7D%5Cfrac%7Bft%7D%7Bs%7D)
Hence, this is the value needed for the second stone to reach the same maximum height than the first stone.