Question

a ball is thrown upward from the ground with an initial velocity of 20 m/s. its height h (in meters) after t seconds is given by h = -5t² + 20t. when does the ball hit the ground? t = 1, t = 0, t = 4, t = 2

Explanation:

Step1: Set height equal to 0

When the ball hits the ground, $h = 0$. So we set up the equation $-5t^{2}+20t = 0$.

Step2: Factor out the common - factor

Factor out $- 5t$ from the left - hand side of the equation: $-5t(t - 4)=0$.

Step3: Use the zero - product property

The zero - product property states that if $ab = 0$, then either $a = 0$ or $b = 0$. So, $-5t=0$ or $t - 4=0$.
Solving $-5t = 0$ gives $t = 0$ (this is the time when the ball is initially thrown).
Solving $t - 4=0$ gives $t = 4$.

Answer:

$t = 4$