How do you find the answer for h(t)=c-(d-4t)^2？?

h(t)=c-(d-4t)^2~ At time t=0, a ball was throuwn 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 above, 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?

How do you find the answer for h(t)=c-(d-4t)^2？?
70 feet
Reply:Since c and d are constants, you need to find their values first before you can find the height of the ball at t = 1. Start by plugging in what you know for t = 2.5.





106 = c - (d - 4(2.5))^2


106 = c - (d - 10)^2


106 = c - (d^2 - 10d + 100)


106 = c - d^2 +10d - 100


206 = c - d^2 + 10d





You know another pair of values for this equation. You know the initial height of the ball (6) at t = 0. Plug in those values and see what you get.





6 = c - (d - 4(0))^2


6 = c - d^2





c - d^2 was in the first equation solved above. Now that you know it's value is always 6 (since c and d are constants), you can substitute 6 for c - d^2.





206 = 6 + 10d


200 = 10d


20 = d





Now that you know the value of d, substitute it into the previous equation to find the value of c.





6 = c - d^2


6 = c - 20^2


6 = c - 400


406 = c





At this point, I like to double check that what I've done works. So, going back to the question of t - 2.5, I'm going to substitute c and d and solve. If I end up with a true statement, I know that I have the values of c and d correct.





206 = c - d^2 + 10d


206 = 406 - 20^2 + 10(20)


206 = 406 - 400 + 200


206 = 6 + 200





Yes! They are correct. Finally, we can get to answering the question. Replace c, d, and t in the question for t = 1 and solve.





h(1) = 406 - (20 - 4(1))^2


h(1) = 406 - (20 - 4) ^2


h(1) = 406 - 16^2


h(1) = 406 - 256


h(1) = 150





The height at t = 1 was 150 ft.