Question

The equation y 4.9t 2 3.5t 2.4 relates the height y in meters to the elapsed time t in seconds for a ball thrown downward at 3.5 meters per second from a height of 2.4 meters from the ground In how many seconds will the ball hit the ground Express your answer as a decimal rounded to the nearest hundredth

Answer 1

Answer:

$t \approx 0.43\,s$

Step-by-step explanation:

The vertical displacement function is $y(t) = -4.9\cdot t^{2}-3.5\cdot t + 2.4$, where $y(t)$ is measured in meters and $t$ in seconds. Ball hits the ground when $y(t) = 0$. That is:

$-4.9\cdot t^{2}-3.5\cdot t + 2.4 = 0$

Whose roots can be found by using the General Formula for Second-Order Polynomials:

$t_{1,2} = \frac{3.5\pm \sqrt{(-3.5)^{2}-4\cdot (-4.9)\cdot (2.4)} }{2\cdot (-4.9)}$

Solutions of this polynomial are:

$t_{1} \approx 0.43\,s,t_{2} \approx -1.14\,s$

Only the first root is physically consistent.