Hi! Can you please explain number 25 section 4 test 1?

Yep! You’re given this equation: h=-4.9t^2+25t. That’s the height of a ball t seconds after it is launched. You need to know how much time goes by before the ball hits the ground. The key thing to remember is that h will equal zero when the ball hits the ground, so you’re really solving 0=-4.9t^2+25t for t.

Since this is the calculator section, you have lots of options at your disposal.

First, you can backsolve. Just plug each answer choice in for t until you get something close to zero as an answer. This is a little tricky because you won’t get exactly zero, but the question does say “approximately.” Below is my calculator’s screen after I tried choice C, recognized that that was probably too high, remembered that a falling ball will get closer and closer to 0 the longer it’s in the air, and then tried D, which turned out to be pretty close to zero.

t1s4-25a

Another way to go is to graph. This is my favorite way for this question, because you can just eyeball the graph and see the right answer—you don’t even need to use the zero function to get it exact. Look:

t1s4-25b

Pretty obvious that that parabola hits y = 0 around x = 5, right?

The last way you can go is to factor and solve algebraically.

0=-4.9t^2+25t

0=t(-4.9t+25)

Since you aren’t interested in the fact that the height was zero at time zero, solve 0=-4.9t+25.

0=-4.9t+25

-25=-4.9t

5.102...=t

Leave a Reply