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 in feet isgiven by the equation s = -2.7t^2 + 50t + 6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b.a. After how many seconds is the ball 20 ft above the moon's surface?After (0.27 , 18.24) seconds the ball will be 20 ft above the moon's surface.(Round to the nearest hundredth as needed. Use a comma to separate answers as needed.)b. How many seconds will it take for the ball to hit the moon’s surface?It will take ___ seconds for the ball to hit the moon’s surface. (Round to the nearest hundredth as needed.)Part A is answered so I just need help with Part B please.

Answer 1

We will have the following:

We are given that the position of the baseball is described by:

`s=-2.7t^2+50t+6.5`

Now, to determine after how much time the baseball hits the Moon's surface we proceed:

`\begin{gathered} -2.7t^2+50t+6.5=0\Rightarrow t=(-(50)\pm√((50)^2-4(-2.7)(6.5)))/(2(-2.7)) \\ \\ x\approx18.65s \\ x\approx-0.13s \end{gathered}`

So, since it would certainly make little sense to talk about negative time in this problem we will have that it would take approximately 18.65 seconds for the baseball to hit the Moon's surface.