Question

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, when the ball will reach a height of 20 feet? Round your answer to two decimal places.

Answer 1

The first time when the ball will reach a height of 20 feet is 0.42 seconds

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

`h = -16t^2 + 40t + 6`

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

`20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0`

Solve by quadractic formula

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}`

`x=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=8,\:b=-20,\:c=7`

`t = (-\left(-20\right)\pm √(\left(-20\right)^2-4\cdot \:8\cdot \:7))/(2\cdot \:8)\\\\t = (20 \pm √(176))/(16)\\\\t = (20 \pm 4√(11))/(16)\\\\t = ( 5 \pm √(11))/(4)\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084`

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds