[SOLVED]initial ground velocity

karush

Well-known member
With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approximately $550$ft.)

well since $f(t)=-16t^2+V_0 t+S_0$ then $550=f(t)$

also, that $f'(0)=V_0$ and $f''(t)=-32$

this is supposed to be solved by using integration but I don't see how this is done without knowing what $t$ is. any suggest?

answer is $187.617\frac{ft}{sec}$

MarkFL

Staff member
With $\displaystyle S_0=0$ we have:

$\displaystyle f(t)=-16t^2+v_0t$ and so:

$\displaystyle f'(t)=-32t+v_0$

What values do the function and its derivative have when the object has reached its desired maximum height?

karush

Well-known member
well the object has $0$ velocity when it reaches max height and since $f'$ is the velocity
we set $0=-32t+V_0$ or $V_0=32t$ but still we have $t$ in this so am going the right direction,

MarkFL

Staff member
Yes, you are doing well, now what is the value of $\displaystyle f(t)$ when the object reaches the maximum height?

karush

Well-known member
Yes, you are doing well, now what is the value of $\displaystyle f(t)$ when the object reaches the maximum height?
max height is given at $550ft$ so $550=-16t^2+(32t)t=32t$ so then $t=5.86$

don't see how this helps if this is plugged back in anywhere

MarkFL

Staff member
This gives you:

$\displaystyle -16t^2+32t^2=16t^2=550$

Solve this for t, then use this value in the formula you found for initial velocity as a function of t.

I know you are to solve this dynamically, but you could also use energy considerations, i.e, equate the initial kinetic energy to the final gravitational potential energy:

$\displaystyle \frac{1}{2}mv_0^2=mgh$

$\displaystyle v_0=\sqrt{32\cdot550}=40\sqrt{22}\,\frac{\text{ft}}{\text{s}}$

karush

Well-known member
appreciate the help, now I see what is happening

well from the TI-nspire I got

$solve(-16t+vt=550,v) v=187.62 ft/s$ where $t=5.86$

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