Question

if someone throws a ball down from a height of 100 meters, the ball's distance from the ground can be modeled by the equation: d = − 9.8 t 2 − 15 t + 100 where t is the time in seconds and d is the distance in meters. At what time (t) will the ball hit the ground? (Hint: what does this mean for d=distance?) You will get two answers. Do both make sense? (Explain in detail Why or Why not

Answer 1

Answer:

The ball will hit the ground when t = 2.52s. The other answer, which is t = -4.05s, does not make sense, because the answer is an instant of time, and there are no negative time measures.

Step-by-step explanation:

To solve this question, we have to find the roots of a quadratic equations, which is explained next.

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem, we have that:

Ball's distance from the ground.

$d(t) = -9.8t^{2} - 15t + 100$

So

$a = -9.8, b = -15, c = 100$

At what time (t) will the ball hit the ground?

This is t when $d(t) = 0$. So

$\bigtriangleup = (-15)^{2} - 4*(-9.8)*100 = 4145$

$t_{1} = \frac{-(-15) + \sqrt{4145}}{2*(-9.8)} = -4.05$

$t_{2} = \frac{-(-15) - \sqrt{4145}}{2*(-9.8)} = 2.52$

The ball will hit the ground when t = 2.52s. The other answer, which is t = -4.05s, does not make sense, because the answer is an instant of time, and there are no negative time measures.