Question

A ball is thrown in the air from the ledge. It’s height in feet is represented by
f(x)= -16(x^2-6x-7), where X is the number of seconds since the ball has been thrown. The height of the ball is 0 feet when it hits the ground. how many seconds does the ball take to reach the ground?

Answer 1

Given ,A ball is thrown in the air from the ledge. It’s height in feet is represented by

A ball is thrown in the air from the ledge. It’s height in feet is represented by f(x)= -16(x^2-6x-7), where X is the number of seconds since the ball has been thrown. The height of the ball is 0 feet when it hits the ground.

$f(x) = 0 \\</p><p>( The \:height \:of \:the \:ball \:is \:0 \:feet \:when \\it \:hits \:the \: ground )$

$\implies -16(x^{2}-6x-7) = 0$

$\implies x^{2} - 6x - 7 = 0$

/* Splitting the middle term,we get */

$\implies x^{2} - 7x + 1x - 7 = 0$

$\implies x(x-7)+1(x-7) = 0$

$\implies (x-7)(x+1) = 0$

$\implies x - 7 = 0 \:Or \:x + 1 = 0$

$\implies x = 7 \:Or \:x = - 1$

/* x should not be negative */

$In \: 7 \:seconds \: the \:ball \:reaches \: the\\ground$

Therefore.,

$\red {Time \:taken \:to \:reach \:the \: ground } \green {= 7 \:seconds }$

•••♪

Answer 2

Answer:

6

Step-by-step explanation: