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 30 ft per sec. The height s of the ball in feet is given by the equation s equals negative 2.7 t squared plus 30 t plus 6.5​, where t is the number of seconds after the ball was thrown. Complete parts a and b.

Answer 1

Answer: 89.803 ft

Explanation:

The complete question is written below:

An astronaut on the moon throws a baseball upward. The astronaut is 6ft, 6 in. tall and the initial velocity of the ball is 30 ft per second. The height s of the ball in feet is given by the equation,s=-2.7t^2+30t+6.5, where t is the number of seconds after the ball is thrown.

The ball will never reach a height of 100ft. How can this be determined algebraically?

We have the following equation that expresses the height $s$ as a function of time:

$s=-2.7t^{2}+30t+6.5$ (1)

Now, if we wan to find the maximum height the baseball reaches and prove it is less than 100 ft, we firstly have to find the time $t_{total}$ the whole parabolic movement lasts and then find $t_{smax}=\frac{t_{total}}{2}$ which is the time it takes the baseball to reach its maximum height.

So, if we want to calculate $t_{total}$, this is fulfilled when $s=0$, when the baseball hits the ground:

$0=-2.7t^{2}+30t+6.5$ (2)

This is a quadratic equation of the form $0=at^{2}+bt+c$, and we have to use the quadratic formula if we want to find  $t_{total}$:

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$  (3)

Where $a=-2.7$, $b=30$, $c=6.5$

Substituting the known values and choosing the positive result of the equation:

$t_{total}=11.323 s$  (4)

Then we can calculate $t_{smax}$:

$t_{smax}=\frac{t_{total}}{2}$ (5)

$t_{smax}=\frac{11.323 s}{2}$

$t_{smax}=5.661 s$  (6)

Substituting (6) in (1):

$s=-2.7(5.661 s)^{2}+30(5.661 s)+6.5$ (7)

$s=89.803 ft$ (8) This is the maximum height the baseball reaches, as we can see it is less than 100 ft