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?

Question

anonymous · Answer

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

anonymous · Answer

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?

anonymous · Answer

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

s · Answer

thank you