An object launched vertically from a point that is 37.5 feet above ground level with an initial velocity of 48.6 feet per second can be represented by the equation h=−16t^2+48.6t+37.5, where h is the height of the object and t is the time after the object is launched. How long does it take the object to hit the ground?

Question

contestada

Respuesta :

gomezgerman032 gomezgerman032 · Answer 1 · 2020-06-30T02:29:16+02:00

Answer: 3.675 seconds

Step-by-step explanation:

Hi, when the object hits the ground, h=0:

h=−16t^2+48.6t+37.5

0=−16t^2+48.6t+37.5

We have to apply the quadratic formula:

For: ax2+ bx + c

x =[ -b ± √b²-4ac] /2a

Replacing with the values given:

a=-16 ; b=48.6; c=37.5

x =[ -(48.6) ± √(-48.6)²-4(-16)37.5] /2(-16)

x = [ -48.6 ± √ 4,761.96] /-32

x = [ -48.6 ± 69] /-32

Positive:

x = [ -48.6 + 69] /-32 = -0.6375

Negative:

x = [ -48.6 - 69] /-32 = 3.675 seconds (seconds can't be negative)

Feel free to ask for more if needed or if you did not understand something.