Question

An object is launched at 19.6 meter per second (m/s) from a 58.8-meter tall platform the equation for the objects height h at the time t seconds after launch is h(t)=4.9^2+19.6t+58.8, where h is in meters. When does the object strike the ground?

Answer 1

h(t)=4.9^2+19.6t+58.8

0=4.9^2+19.6t+58.8

4.9^2+19.6t+58.8 = 0

Now use the quadratic formula to solve for t

Answer 2

If you forget the formula, it is

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 3

thank you so much, but instead of using the quadratic formula couldn't i just factor that part out.

Answer 4

thank you so much, but instead of using the quadratic formula couldn't i just factor that part out. @jim_thompson5910

Answer 5

you could, but it's a lot more work and factoring doesn't always work

Answer 6

can you also help me with this one? The length of the rectangle is 6in more than its width and the area of the rectangle is 91in^2. find dimensions of the rectangle.

Answer 7

you have to do this if you want to factor


4.9t^2+19.6t+58.8 = 0

10(4.9t^2+19.6t+58.8) = 10*0

49t^2 + 196t + 588 = 0

then you have to find two numbers that multiply to 49*588 = 28,812 and add to 196...no easy task

Answer 8

okay

Answer 9

yeah, that's why the quadratic formula is much better

Answer 10

i did the Q formula and i don't know if i did it right once i get down to the end how solve do i just divide and get 4. @jim_thompson5910

Answer 11

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\Large x=\frac{-19.6 \pm \sqrt{(19.6)^2-4(-4.9)(58.8)}}{2*(-4.9)}\]

\[\Large x=\frac{-19.6 \pm \sqrt{1536.64}}{-9.8}\]

\[\Large x=\frac{-19.6 \pm 39.2}{-9.8}\]

\[\Large x=\frac{-19.6 + 39.2}{-9.8} \ \text{or} \ x = \frac{-19.6 - 39.2}{-9.8}\]

\[\Large x= -2\ \text{or} \ x = 6\]

Answer 12

ignore the negative solution to get the only solution to be x = 6

so it will take 6 seconds for the object to hit the ground

Answer 13

The length of the rectangle is 6in more than its width and the area of the rectangle is 91in^2. find dimensions of the rectangle. @jim_thompson5910

Answer 14

Height equals zero at the instant the object strikes the ground.

Answer 15

"The length of the rectangle is 6in more than its width" ---> L = W + 6

Area:

A = LW

A = (W+6)*W

A = W(W+6)

A = x(x+6)

91 = x^2 + 6x

x^2 + 6x - 91 = 0

Now use the quadratic formula to solve for x. Optionally you can factor if you want here.

Answer 16

@jim_thompson5910 i tried to factor, but i couldn't find where they meet at

Answer 17

hint: 13 and -7 multiply to -91 and add to 6

Answer 18

@jim_thompson5910 thank you for alll your help (:

Answer 19

yw