## S 4 years ago An object is thrown up from a height of 160 feet. Its velocity at t is v(t)=-32t+24 ft/ sec a) how high will it go? b) with what velocity will it hit the ground?

1. anonymous

max height is at the vertex. use $$t=-\frac{b}{2a}$$ to find it

2. anonymous

oh but first you need the equation. it is $h(t)=-16t^2+v_0t+h_0$ in your case $$h_0=160$$ and $$v_0=$$ hmm confused. what is the initial velocity?

3. anonymous

oh it is 24, so your equation is $h(t)=-16t^2+24t+160$ max when $$t=\frac{24}{32}=\frac{3}{4}$$ replace $$t$$ by $$\frac{3}{4}$$ to find the max height

4. anonymous

thank you