Question

A stone is thrown with an initial velocity of 30 ​ft/s from the edge of a bridge that is 43 ft above the ground. The height of this stone above the ground t seconds after it is thrown is ​f(t)equalsminus16t squaredplus30tplus43. If a second stone is thrown from the​ ground, then its height above the ground after t seconds is given by ​g(t)equalsminus16t squaredplusv0​t, where v0 is the initial velocity of the second stone. Determine the value of v0 such that the two stones reach the same high point.

Answer 1

Answer:

The value of $v_{0}$ such that the two stones reaches the same maximum height is

$v_{0}=\frac{\sqrt{913} }{2} \frac{ft}{s}$

Explanation:

For the first stone, we need to find the maximum height reached, and for that we have to derivate the given position function

$f(t)=-16t^{2}+30t+43$

derivating, we get

$f'(t)=-32t+30$

now we have to equalize the derivate to zero, and clear t

$0=-32t+30$

$t=\frac{30}{32}s$

then, if we put this value of t in the position function, we obtain that the maximum height for the stone is

$f_{max}=\frac{913}{16}ft$

For the other stone, we have the given position function

$f(t)=-16t^{2}+v_{0}t$

And again, derivating and clearing t, we obtain

$t=\frac{v_{0}}{32}s$

As the height must be the same for both stones, we can substitute in the position function theese values

$\frac{913}{16}=-16(\frac{v_{0}}{32})^{2}+v_{0}(\frac{v_{0}}{32})$

From where our value for $v_{0}$ results to be

$v_{0}=\frac{\sqrt{913} }{2}\frac{ft}{s}$

Hence, this is the value needed for the second stone to reach the same maximum height than the first stone.