Suppose that the position y (feet) of a projectile above the ground is described by the equation y=-16t^2+70t+2, where t is the time elapsed (seconds) since the object was launched. (A) What is the initial position of the object? (B) When will the object strike the ground? Please help!!!! :(
(A) when t=0 , y = 2
(B) the ground is y=0
... 0 = -16t^2 + 70t + 2 ... solve the quadratic for t to find the impact time
... use the quadratic formula
To find the initial position of the object, we need to find the value of y when t is equal to 0. This is because at t=0, the object is just launched and has not yet traveled any distance.
So to find the initial position:
A) Plug in t=0 into the equation y = -16t^2 + 70t + 2:
y = -16(0)^2 + 70(0) + 2
y = 0 + 0 + 2
y = 2
Therefore, the initial position of the object is 2 feet.
Now let's move on to the second part of the question:
B) To determine when the object will strike the ground, we need to find the value(s) of t when y equals 0. This is because when y is 0, the object is at ground level.
So we set y = 0 in the equation and solve for t:
0 = -16t^2 + 70t + 2
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 70, and c = 2. Plugging these values into the quadratic formula:
t = (-(70) ± √((70)^2 - 4(-16)(2))) / (2(-16))
t = (-70 ± √(4900 + 128)) / (-32)
t = (-70 ± √(5028)) / (-32)
To simplify the expression, we can find the square root of 5028, which is approximately 70.99.
t = (-70 ± 70.99) / (-32)
Now we have two possible values for t:
t1 = (-70 + 70.99) / (-32)
t2 = (-70 - 70.99) / (-32)
Evaluating these values, we get:
t1 ≈ 0.03 seconds
t2 ≈ 4.42 seconds
Therefore, the object will strike the ground approximately 0.03 seconds and 4.42 seconds after it was launched.