A cannonball is shot into the air in the equation as shown below. The height of the ball above the ground, in metres, t seconds after being shot is approximated by y=-5t2+20t+160. Determine after how many seconds the cannon reaches its maximum height and what is the maximum height? Solve by completing the square or factoring/axis/vertex.
To find the maximum height of the cannonball, we need to determine the value of t at which the equation y = -5t^2 + 20t + 160 reaches its highest point.
One way to find this is by completing the square. We rewrite the equation as follows:
y = -5(t^2 - 4t) + 160
Next, we need to complete the square inside the parentheses. To do so, we take half of the coefficient of t (-4) and square it:
(-4/2)^2 = (-2)^2 = 4.
Now, we can add and subtract this value (4) inside the parentheses:
y = -5(t^2 - 4t + 4 - 4) + 160
= -5((t - 2)^2 - 4) + 160.
Simplifying further, we have:
y = -5(t - 2)^2 + 20 + 160
= -5(t - 2)^2 + 180.
Now, we can see that the equation is in the form y = a(x - h)^2 + k, where (h, k) is the coordinates of the vertex. In this case, the vertex is given by (2, 180). Therefore, the maximum height of the cannonball is 180 meters and it is reached after 2 seconds.
So, after 2 seconds, the cannon reaches its maximum height, which is 180 meters.
To determine the time at which the cannon reaches its maximum height, we can look at the equation y = -5t^2 + 20t + 160 and find the vertex of the parabolic function. The vertex represents the maximum point of the parabola, i.e., the highest point the cannonball reaches.
To find the time at which the maximum height occurs, we need to find the x-coordinate of the vertex. This can be done by using the formula x = -b / (2a), where a, b, and c are the coefficients in the quadratic equation.
In this case, the quadratic equation is y = -5t^2 + 20t + 160, where a = -5 and b = 20. Plugging these values into the formula, we get:
t = -20 / (2 * -5)
= -20 / -10
= 2
So, after 2 seconds, the cannon reaches its maximum height.
Now, to find the maximum height, we substitute this value of t back into the equation y = -5t^2 + 20t + 160:
y = -5(2)^2 + 20(2) + 160
= -5(4) + 40 + 160
= -20 + 40 + 160
= 180
Therefore, the maximum height reached by the cannonball is 180 meters.
So, after 2 seconds, the cannon reaches its maximum height of 180 meters.