A baseball player hits a baseball into the air with an initial vertical velocity of 48 feet per second from a height of 3 feet.

a) Write an equation that gives the baseball's height as a function of the time (in seconds) after it is hit.

b) After how many seconds does the baseball reach its maximum height?

c) What is the maximum height?

Question

A baseball player hits a baseball into the air with an initial vertical velocity of 48 feet per second from a height of 3 feet.

a) Write an equation that gives the baseball's height as a function of the time (in seconds) after it is hit.

b) After how many seconds does the baseball reach its maximum height?

c) What is the maximum height?

anonymous · Answer

$$y = v _{0} t \times \sin a - \frac{ g t ^{2} }{ 2 }$$ maybe that helps

anonymous · Answer

a.) Distance in the vertical= vinitial*time+1/2(acceleration due to gravity*(time)^2)
b.)Take derivative and set = 0, then solve for t.
c.)plug value for t in a.) and solve for distance in the vertical.

anonymous · Answer

btw your only given a vertical velocity, so voila!

anonymous · Answer

accleration due to gravity (g) is a constant, 9.8m/s^2. The baseball acts opposite gravity therefore it has a negative acceleration.

anonymous · Answer

is this right for a)? $$h=-16t^{2}+48t+3$$

jim766 · Answer

yeah I think so...

jim766 · Answer

do you know how to find time till max hieight?

anonymous · Answer

no not really

jim766 · Answer

H = Ax^2 + Bx + C       is standard from
A = -16
B = 48
C = 3                   

max will be at   -B/2A

anonymous · Answer

so would 1.5 be right?

jim766 · Answer

Thats good....  that is how many seconds before the ball hits max

jim766 · Answer

Now to find the max height , substitute   1.5 for x in the equation

anonymous · Answer

So, 39 feet would be the max height?

jim766 · Answer

perfect!

anonymous · Answer

Thank you so much... You are a life saver!

jim766 · Answer

yw