An object is fired vertically upward with an initial speed of 68.6 meters per second. After t seconds, the height is shown by y = 4.9t^ + 68.6 t meters. Given that this object has an initial speed of 68.6 m/s, what is the maximum height it will reach?
if height y is
y= 4.9t^2+ 68.6t, it is a very odd object, gravity is not going to ever pull it downward. I suggest this is more likely in the real world:
y= - 4.9t^2+68.6t
so y max can be found when
y'=0=-9.8t+68.6 and solve that for t, the time at the max.
then put that t into the y equation.
To find the maximum height reached by the object, we need to find the vertex of the quadratic function y(t) = 4.9t^2 + 68.6t.
The vertex of a quadratic function in the form y = ax^2 + bx + c is given by the formula t = -b / (2a).
In this case, a = 4.9 and b = 68.6.
t = -68.6 / (2 * 4.9)
t = -6.98 seconds
To find the maximum height, we substitute this value of t back into the equation:
y = 4.9t^2 + 68.6t
y = 4.9(-6.98)^2 + 68.6(-6.98)
y ≈ -242.36 meters
Therefore, the maximum height reached by the object is approximately 242.36 meters.
To find the maximum height the object will reach, we need to determine the vertex of the parabolic equation y = 4.9t² + 68.6t.
The vertex of a parabola in the form y = ax² + bx + c can be found using the formula t = -b / (2a).
In this case, a = 4.9 and b = 68.6. Substituting these values into the formula, we get:
t = -68.6 / (2 * 4.9)
t = -68.6 / 9.8
t ≈ -7
Since time cannot be negative, we can ignore the negative value and consider t = 7 seconds as the time when the object reaches its maximum height.
To find the corresponding height, we substitute t = 7 into the equation:
y = 4.9(7)² + 68.6(7)
y = 4.9(49) + 480.2
y = 240.1 + 480.2
y = 720.3
Therefore, the maximum height the object will reach is approximately 720.3 meters.