Mitch throws a baseball straight up in the air from a cliff that's 48 ft high. The initial velocity is 48 ft per second.

Question

Mitch throws a baseball straight up in the air from a cliff that's 48 ft high. The initial velocity is 48 ft per second. The height (in feet)of the object after t second is given by h(t) = -16t^2 + 48t +48. How high will the ball be after 2 seconds? How long will it take to reach maximum height? And find the time at which the height of the object is 64 ft (round decimal place)

Answer 1

(a) h(t) gives the height of the baseball at time t. So to find the height h after 2 seconds, set t = 2 in the equation for h(t) and solve:

 

h(2) = -16(2)² + 48(2) + 48 = 80 feet.

 

(b) The maximum height is reached when the velocity of the baseball is 0. In order to find an equation for the velocity, differentiate the position function with respect to t:

 

velocity = v(t) = h'(t) = -32t  +48.

 

We want to know the time at which h'(t) = 0, so set h'(t) = 0 and solve for t:

 

-32t + 48 = 0     or      -16(2t-3) = 0,

 

and so t = 3/2 seconds. Thus the ball reaches its max height at 1.5 seconds after launch.

 

(c) To find the time at which the ball is at a height of 64 feet, set h(t) = 64 and solve for t:

 

-16t² + 48t + 48 = 64, or -16(t² -3t + 1) = 0.

 

Use the quadratic formula to yield two solutions for t: t = (3 ± √5)/2, which in decimal form is approximately 

 

t = .38 and t = 2.6 seconds. 

 

So there are two times at which the ball reaches a height of 64 feet: once on the way up and once on the way down.