A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 60 feet for second. How high will the ball go? a(t)=-32 feet per second is the acceleration of gravity.Problem involves integrals.

Question

anonymous · Answer

$$h(t)=-16t^2+v_0t+h_0$$in your case you have $v_0=60, h_0=6$ i guess giving you the equation 
$$h(t)=-16t^2+60t+6$$ max is at the vertex, where $t=-\frac{b}{2a}=-\frac{60}{2\times -16}=\frac{15}{8}$

anonymous · Answer

replace $t$ by $\frac{15}{8}$ to get the maximum height

anonymous · Answer

you got this?

anonymous · Answer

where did you get the -16 from?

anonymous · Answer

aah good question 
first of because i know it.  second because the acceleration is constatn at -32
this is the second derivative of the position function
so first derivative is 
$$-32t+v_0$$ and position funciton itself if 
$$-16t^2+v_0t+h_0$$ by taking anti derivatives twice

anonymous · Answer

i should say acceleration due to gravity is -32 feet per second squared

anonymous · Answer

ohhhok..got it! thanks!!

anonymous · Answer

yw