Quadratic equations

Question

an object is launched 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t)=-4.9t²+19.6t +58.8, where s is in meters.

when does the object strike the ground?

Answer 1

when object strikes the ground, it's height from the ground would be zero, so you need to solve the quadratic:

 

s(t)=-4.9t² + 19.6t + 58.8 = 0, to get the value of t

 

you can either apply the quadratic formula right away: t = (-b ± √b² -4ac ) / 2a

 

Or, what I prefer is to simplify the quadratic before applying the formula.

 

-4.9t² + 19.6t + 58.8 = 0

=> 49t² - 196t - 588 = 0 (multiplied both sides by -10)

=> 7t² - 28t - 84 = 0 (divided by 7)

=> t² - 4t - 12 = 0 (divided by 7 again)

Now, it's much easier to apply the formula:

t = (4 ± √ (-4)² -4*1*(-12) ) / 2*1

t = (4 ± √ 16 + 48 ) / 2

t = (4 ± 8 ) / 2

t = 6 or -2 seconds

since time can not be negative, your answer is 6 seconds