When a ball is thrown upward, its height, h, in metres, is given by h=1.5 +19.6t -4.9t², where t is the number of seconds after it is thrown. For what length of time is the ball above the ground?
the ball is above the ground when h > 0
1.5 + 19.6t - 4.9t^2 > 0
49t^2 - 196t - 15 < 0
let's solve it as an equation.....
t = (196 ± √41356)/98 = 4.075 or a negative t
but t ≥ 0
so the ball is above the ground between
t=0 and t=4.075
that is, it was above ground for 4.075 seconds
To find the length of time the ball is above the ground, we need to determine the values of 't' when the height 'h' is greater than zero.
Given that the ball's height is given by the equation h = 1.5 + 19.6t - 4.9t², we can set this expression equal to zero to find the time at which the ball hits the ground.
0 = 1.5 + 19.6t - 4.9t²
Rearranging the equation:
4.9t² - 19.6t - 1.5 = 0
We can solve this quadratic equation to find the values of 't' when the ball is on the ground or above it.
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
We substitute a = 4.9, b = -19.6, and c = -1.5 into the equation:
t = (19.6 ± √((-19.6)² - 4 * 4.9 * -1.5)) / (2 * 4.9)
After calculating this expression, we will have two values of 't'. The smaller one corresponds to the moment the ball hits the ground, and the larger one represents the time when the ball is thrown upward and subsequently falls back to the ground.
Therefore, the length of time the ball is above the ground is the difference between the larger value 't' and the smaller value 't'.