<p>can someone help me with this problem? thanks.</p>
<p>at time t=0, a ball was thrown upward from an initial height of 6 feet. until the ball hit the ground, its height, in feet, after t seconds was given by the function h(t)=c-(d-4t)^2, in which c and d are positive constants. if the ball reached its maximum height of 106 feet at time t=2.5, what was the height, in feet, of the ball at time t=1?</p>
<p>Use calc. The derivation at t=2.5 must be zero since thats when the ball reached its maximum height. Plug in what you get for d to find c. Then just find h(1).</p>
<p>Too lazy to do it out. If you need more help just ask.</p>
<p>Mathwiz said it.
h(t)=c-(d-4t)^2
dh/dt h(t) = -2(d-4t)(4)=0 therefore d-4t=0 and t=2.5 so d=10
at h(0) = 6 = c-(10)^2 and so c = 106 and we know that d=10
h(1) = 106 - (10-4(1))^2 = 106-36 = 70ft</p>