Richard tosses a ball into the air. The funtion h(t)=-5t

Question

Richard tosses a ball into the air. The funtion h(t)=-5t² +10t+6 gives the approximate height h In meters, of the ball t seconds after he tosses it. Dose the ball reach a height of 12 m?

Answer 1

There are two easy ways to find the vertex of a parabola. Graphing it is one way, either on your calculator or on desmos.com. Just find the highest (or lowest) point, and the (x,y) coordinates should be easy to find.

The other way is, given a quadratic in standard form, ax²+bx+c, the x-coordinate of the vertex is given by -b/2a. Since your quadratic is in standard form, we just need to identify a and b.

Given h(t)=-5t² +10t+6, we see the coefficient of the t² term is -5. a=-5. The next coefficient is b=10. Therefore, -b/2a= -10/(2(-5))= -10/-10 = 1. That's the x-coordinate, or in this case, t. Since t=1, the ball reaches its highest point one second after its thrown. To find it's height, plug 1 in for t: h(1)=-5(1)² +10(1)+6 = -5+10+6 = 11.

The ball only reaches 11 meters, so it doesn't make it to 12.

I hope this helps! Let me know if there's something you don't understand?

Have fun mathing!

LizZ

Answer 2

One thing you can try: the Client wants to know, can h(t) ever >= 12 m?

So set h(t) = 12 and see what time values "t" can produce a height of 12 m, if any.

Estimating, note that at t = 1, h(1) = 11 m. So, is the ball still going up after 1 second? hmm.