math hw help needed

Question

An object is thrown upward at a speed of 75 feet per second by a machine from a height of 19 feet off the ground. The height h of the object after t seconds can be found using the equation h=−16t^2+75t+19h

When will the height be 33 feet?

 Select an answer seconds per foot, seconds,  feet, feet per second 

When will the object reach the ground?

 Select an answer feet per second, seconds per foot, seconds , feet 

Answer 1

The equation for the height of the object after t seconds should be:

h(t) = -16t² + 75t +19

To find t when the height of the object as at 33 ft, set h = 33 and solve:

33 = -16t² + 75t + 19

Bring all terms over to the left side, remembering to change signs:

16t² -75t +(33-19) = 16t² -75t + 14 = 0

You now need to solve a quadratic equation in t. (Note WHEN means time, so the solution t should be in seconds)

Since the equation

16t² - 75t + 14 = 0 (which is of the form ax² + bx + c = 0)

cannot be factored, you will need to solve it using the quadratic formula:

t = [-b + sqrt (b²- 4*a*c)]/2*a

where plugging in: a = 16, b= -75, and c = 14 will give you

t = [75 + sqrt ((-75)²-4(16)(14))] / 2(16)

t = 0.195 s and 4.493 s, which we may round to 0.2 seconds and 4.5 seconds

This means that the object will reach 33 ft twice, once when it is going up (remember that it started already at 19 feet, so it only needed to go up 14 feet), and then when it is coming back down.

To solve for when the object will reach the ground, you just set h = 0, giving you the equation 0 = -16t² + 75t + 19, but moving everything over to the left and changing signs, you get 16t²- 75t -19 = 0 which you will need to solve using the quadratic formula again.

when you plug in a = 16, b = -75 , c =-19

t = [75 + sqrt ((-75)²-4(16)(-19))] / 2(16)

you get t = 4.93 seconds (the other solution for t is negative, which isn't possible)

Answer 2

h= -16t^2 +75t + 19

33 = -16t^2 +75t +19

16t^2 -75t + 14 = 0

use the quadratic formula t=-b/2a+/- (1/2a)sqr(b^2-4ac)

t= 75/32 + or - (1/32)sqr(5625 -896)

t= (75 +/- 68.77)/32

t= 6.23/32 or 143.77/32

t = 0.19 or 4.49 seconds when height = 33 feet

about 1/5 or 4 1/2 seconds

it hits the ground when

h=0 = -16t^2 +75t +19

16t^2 -75t -19=0

t = 75/32 +/- (1/32)sqr(5625+1216)

t= (75+/- 82.71)/32

t = -0.24 or 4.93 seconds to reach ground level

almost 5 seconds

ignore the negative solution