the height of an infield fly ball, h metres, t seconds after it is hit is given by the form h=1.5 + 30t - 4.9t^2

a. determine the maximum height of the ball. i keep gettin -13.5, cud this b rite because it is infield?
b. how long does it take the ball to reach maximum height? 7.5 seconds
c. how long does it take the ball to reach the ground? 15 seconds
d. how high is the ball when it is hit? 1.5 metres

2. Originally Posted by justine_0612
the height of an infield fly ball, h metres, t seconds after it is hit is given by the form h=1.5 + 30t - 4.9t^2

a. determine the maximum height of the ball. i keep gettin -13.5, cud this b rite because it is infield?
b. how long does it take the ball to reach maximum height? 7.5 seconds
c. how long does it take the ball to reach the ground? 15 seconds
d. how high is the ball when it is hit? 1.5 metres
Unfortunately, you have not shared with us your workings.

a. What has 'infield' to do with anything? Please do not try to talk yourself into entirely unreasonble results. The maximum height cannot by UNDERground.

b. WAY too long. How did you arrive at this answer? Hint: -b/2a

c. Again, WAY too long.

d. Got this one.

Perhaps you are confusing the coefficients. The coefficient on $t^{2}$ is 'a' in the standard way of setting things up. Thus a = -4.9 m/ $s^{2}$

3. Hello, Justine!

The height of an infield fly ball, h metres, t seconds after it is hit
is given by the form: . $h \:=\:1.5 + 30t - 4.9t^2$

(a) Determine the maximum height of the ball. .I keep gettin -13.5
We have a down-opening parabola.
. . It is at a maximum at its vertex: . $t \:=\:\frac{-b}{2a}$

We have: . $a = -4.9,\;b = 30,\;c = 1.5$

. . Hence: . $t \:=\:\frac{-30}{2(-4.9)} \:\approx\:3.06$ seconds.

Therefore: . $h \;=\;1.5 + 30(3.06) - 4.9(3.06^2) \:\approx\:\boxed{47.4\text{ meters}}$

(b) How long does it take the ball to reach maximum height? . 7.5 seconds . . . . no
We found this answer is part (a): . $\boxed{3.06\text{ seconds}}$

(c) How long does it take the ball to reach the ground? . 15 seconds . . . . no

It "reaches the ground" when $h = 0.$

We have: . $1.5 + 30t - 4.9t^2 \:=\:0\quad\Rightarrow\quad 49t^2 - 300t - 15 \:=\:0$

Quadratic Formula: . $t \;=\;\frac{300 \pm\sqrt{300^2 - 4(49)(-15)}}{2(49)}$

. . $t \;=\;\frac{300 \pm\sqrt{93940}}{98} \;=\;\frac{300 \pm 304.8606239}{98} \;\approx\;6.17,\;-0.05$

Since $t$ must be positive: . $\boxed{t \:=\:6.17\text{ seconds}}$

(d) How high is the ball when it is hit? . 1.5 metres . . . . Right!