quadratic equation

Question

An astronaut on the moon throws a baseball upward. The astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is 50 ft. per sec. The height s of the ball is given by the equation s=-2.7t²+50t+6.5, where t is the number of seconds after the ball was thrown.

a- After how many seconds is the ball 18 ft above the​ moon's surface?

b- what is the maximum height of the ball

c- how many seconds will it take for the ball to touch the moons surface

Answer 1

Here's a bit of the helpppppppppp you asked for...

s(t) = –2.7t² + 50t + 6.5 is the equation for the height of the ball (in ft) as a function of t (in s).

a- After how many seconds is the ball 18 ft above the​ moon's surface?

Set s(t) = 18 and solve for t.

18 = –2.7t² + 50t + 6.5 leads to the quadratic equation 2.7t² – 50t + 11.5 = 0

which you can solve using the quadratic formula x = (–b ±√(b² – 4ac))/2a

by plugging in the coefficients a = 2.7, b = – 50, and c = 11.5,

a straightforward (albeit very tedious) route to the answer.

The will be two solutions. You want the lower number (which will be just a fraction of a second) telling you when the height of the ball is18 ft on its way up; the second higher number (> 18 s) will tell you when the height of the ball is again 18 ft on its way down.

b- what is the maximum height of the ball?

Go back to the equation s(t) = –2.7t² + 50t + 6.5 which is a quadratic equation

with coefficients a = –2.7, b = 50, and c = 6.5

The time, t, when the ball reaches its maximum height is calculated directly from these coefficients:

t = –b/2a = –50/(–5.4) ≈ 9.26 s

the maximum height is a simple calculation:

s(9.26) = –2.7(9.26)² + 50(9.26) + 6.5

c- how many seconds will it take for the ball to touch the moons surface?

Set s(t) = 0 and solve for t.

–2.7t² + 50t + 6.5 = 0 can be solved using the quadratic formula x = (–b ±√(b² – 4ac))/2a

As in part a, this is a straightforward (albeit very tedious) route to the answer.

There will be two solutions. Here you want throw out the lower number (because it will be negative) and choose the higher one (which will be between 18 and 19 s).

Answer 2

a- Replace s by 18 in the equation and solve for t using thr quadratic formula. Remember t, time cannot be negative. t=(-50+/_ sqrt(50² -4×-2.7×-11.5))/(2×-2.7)

^{b - To find the maximum height you need to find the line of symmetry which is given by the formula t (time) = -b/2a = -50/(2×-2 7). Then replace the value of t in the quadratic equation S, to find the maximum height.}

^{c- replace s by zero and solve for t}

^{0 = -2.7t2+50t+6.5}