Algebra question?

Question

If an object is thrown upward with an initial velocity of 64 ft/sec, its height after t sec is given by h = 64t - 16t2. Find the number of seconds the object is in the air before it hits the ground.

Answer 1

when the oject hits the ground then h=0 put h=0 in given equation and we get 64t-16t^2=0
===>16t(4-t)=0
===>t=0seconds(initial position) or t=4 seconds(after it hits the ground.
So stone was in air for 4 seconds before it hit the ground

Answer 2

solve 0 = 64t + 16t^2 (hitting the ground = hight of zero). This can be factored as

16t(4 - t) = 0

Which describes a parabola with two zeros. If t=0, the left term is zero (16*0) = 0, so the object is at the ground at t=0 (at the instant when you throw it). The other zero is when t=4 (4-4) = 0.

Hence, 4 seconds.

Answer 3

you want to know when the ball hits the ground so this means you need to substitute h = 0 (hence you are finding a value for t when h = 0).
so

64t - 16t^2 = 0 (the ^2 means that it is squared)

then by taking a common factor of 16t we get

16t ( 4 - t ) = 0

so

16t = 0 >> t = 0
4 - t = 0 >> t = 4

t = 0 cannot be the correct answer

the answer is t = 4

=)

Answer 4

-16t^2 +64t = 0
-16t ( t-4)=0
t-4=0
t=4 seconds

Answer 5

h=64t-16t^2

when the object hits the ground, h=0

so 0=64t-16t^2=16t(4-t)

t= 4... number of seconds is 4.

Answer 6

Set h = 0
64t - 16t^2 = 0
Factor
16t(4-t) = 0
16t = 0 or 4-t = 0
t = 0 or t = 4
Therefore, the ball is in the air 4-0 = 4 seconds.

Answer 7

For increasing cubic binomials the final formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^one million + 3*a^one million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^one million + 3*x^one million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Answer 8

h = 64t - 16t2
0=64t-16t^2
16t^2-64t=0
16t(t-4)=0
t=0 start
t-4=0
t=4seconds when it comes back down.