If an object is thrown upward with an initial velocity of 96 feet/sec, it's height after t sec is given by h=96t-16t^2. Find the number of seconds before the object hits the ground.
When it hits the ground, h is again zero
h=0=96t-16t^2=16t(6-t)
so t=6seconds
To find the number of seconds before the object hits the ground, we need to determine when the height (h) becomes zero. In other words, we want to find the value of t that makes the equation h = 96t - 16t^2 equal to zero.
The height equation is already given as h = 96t - 16t^2. To find when h = 0, we set the equation equal to zero:
0 = 96t - 16t^2
Now, we have a quadratic equation. To solve for t, we can factor or use the quadratic formula.
Let's use the quadratic formula:
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, the coefficient of t^2 is -16, the coefficient of t is 96, and the constant term is 0.
Applying the quadratic formula, we have:
t = (-96 ± √(96^2 - 4(-16)(0))) / 2(-16)
Simplifying further:
t = (-96 ± √(9216)) / -32
t = (-96 ± 96) / -32
t = -96/(-32) ± 96/(-32)
t = 3 ± 3
Therefore, we get two possible solutions:
t1 = 3 + 3 = 6
t2 = 3 - 3 = 0 (discard this solution)
Since time (t) cannot be negative, the only valid solution is t = 6.
Therefore, the object hits the ground after 6 seconds.
To find the number of seconds before the object hits the ground, we need to find the time when the height (h) becomes 0.
The equation for the height of the object after t seconds is given by:
h = 96t - 16t^2
Setting h to 0, we can solve for t:
0 = 96t - 16t^2
Rearranging the equation:
16t^2 - 96t = 0
Factoring out a common factor of 16t:
16t(t - 6) = 0
Using the zero product property, either 16t = 0 or t - 6 = 0:
16t = 0 or t - 6 = 0
Solving each equation:
16t = 0 --> t = 0 (This represents the initial time when the object is thrown)
t - 6 = 0 --> t = 6 (This represents the time at which the object hits the ground)
Therefore, the object hits the ground after 6 seconds.