Natalie found a tennis ball outside a tennis court. She picked up the ball and threw it over the fence into the court. The path of the ball can represented by h= -16t (squared) + 18t +5
a. Find the maximum height of the tennis ball.
b. How long will it take to reach the ground?
The equation of motion is,
h = -16t^2 +18t + 5
When the ball starts off, at t = 0, the height of the ball is h = -0 + 0 + 5 = 5
h = 5, at t = 0
The ball will reach a maximum height, and then fall back down to a height of h = 5, in a time taken, of t = 2T, where T is the time to reach its maximum height, after being projected from an inital height of h = 5.
So, in the equation of motion. h = -16t^2 + 18t + 5, we set h = 5 and solve for t, to fond the time it takes to reach max heright and fall back to its initial height of 5 again.
5 = -16t^2 + 18t + 5
16t^2 - 18t = 0
8t^2 - 9t = 0
t(8t - 9) = 0
t = 0 or t = 9/8
But t = 2T
Hence, T = 9/16 s
(a) Max height is given by substituting t = 9/16 in the equation of motion.
h = -16(9/16)^2 + 18(9/16) + 5
h_max = 10.0625
(b) To reach the ground, set h = 0 in the equation of motion
0 = -16t^2 + 18t + 5
16t^2 - 18t - 5 = 0
We now simply solve the quadratic equation for t.
t = {18 +/- sqrt[18^2 - 4*16*(-5)]} / (2*16)
t = {18 +/- sqrt[324 + 320]} / (32)
t = {18 +/- 25.377} / (32)
t = 1.355 s (we ignore the negative solution)