After t seconds, a ball tossed in the air from the ground level reaches a height of h feet given by the equation h = 144t-16t^2
What is the height of the ball after 3 seconds?
What is the maximum height the ball will reach?
Find the number of seconds the ball is in the air when it reaches a height of 224 feet.
After how many seconds will the ball hit the ground before rebound?

144t-16t^2= 16t(9-t)
after 3 seconds (just substitute t=3):
16(3)(9-3)=
16(3)(6)=
16*18=
288
maximum height: find the average of the roots:
the roots of 16t(9-t) are t=0 or t=9
since it’s a parabola, the maximum is at t=4.5, at 324 ft
if its height is 224 feet, the equation is 224=144t-16t^2
16t^2-144t+224=0
divide by 16: t^2-9t+14=0
this can be factored as (t-2)(t-7)=0
the roots are t=2 and t=7, so the ball has been in the air for either 2 seconds or 7 seconds
the roots to 144t-16t^2 are 0 and 9, so the ball will hit the ground 9 seconds after being thrown

H = -16t^2 + 144t
when you substitute a number in for t, it will tell you how high in the air the ball is after that many seconds. to find the height at three seconds, put in 3 for t.
h = -16*9 + 144*3
h = -144 + 3*144
h = 2*144 (i didn’t want to multiply out 3*144 cause im lazy, so i combined like terms as if 144 was a variable. )
h = 288
after 3 seconds, the ball is 288ft in the air.

the maximum height is more complicated (and feel free to skip this paragraph if you think it’ll confuse you more. ) you could do it with a graphing calculator by graphing the function f(x)=-16x^2+144t and using a maximum feature. you could also use differential calculus to find it. but i think the best way is the way suggested by the person who answered before me, that is, finding the average of the roots. since the parabola’s line of symmetry is perpendicular to the x-axis, that means the places the parabola intersects the x-axis (the roots) will be the same distance from the axis of symmetry and therefore the same distance from the vertex (the maximum point of the parabola. )

the roots of the parabola can be found this way:
h = -16t^2 + 144t
0 =  -16t^2 + 144t
0 = t(-16t + 144)
0 = t or 0 = -16t + 144
16t = 144
t = 9
the roots are t=0 and t=9, therefore the line of symmetry (the middle of the parabola) is t=4.5. to find the maximum point of the parabola, plug 4.5 into the function for t and find the resulting height.
h = -16(4.5)^2 + 144(4.5)
h =  -324 + 648
h = 324
so the maximum height happens at t=4.5 seconds, and the maximum height is 324ft.

to find how long it takes for the ball to be at 224 feet in the air, substitute 224 in for h:
h = -16t^2 + 144t
224 = -16t^2 + 144t
56 = -4t^2 + 36t divide by 4
14 = -t^2 + 9t divide by 4 again
t^2 - 9t + 14 = 0 rearrange
(t - 7)(t - 2) = 0
t = 7 and t = 2
at 2 seconds and again at 7 seconds, the ball will be 224ft in the air.

after how many seconds will the ball hit the ground? we already know this. the ball touches the ground just like the parabola touches the x-axis. to find when the ball hits the ground, we just have to find the root of the parabola, and we already know from earlier that the roots are t=0 and t=9. so, the ball will come into contact with the ground at 9 seconds.

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