To find the time at which the ball reaches its maximum height, we need to find the vertex of the parabolic function h(t) = -0.2t + 2t.
The t-coordinate of the vertex of a parabolic function with equation ax^2 + bx + c is given by t = -b / 2a. In this case, a = -0.2 and b = 2. Plugging these values in, we get t = -2 / 2(-0.2) = 5 seconds.
Therefore, the ball reaches its maximum height after 5 seconds.
To find the maximum height of the ball, we substitute t = 5 into the function h(t) = -0.2t + 2t.
h(5) = -0.2(5) + 2(5) = -1 + 10 = 9 feet.
Therefore, the maximum height of the ball is 9 feet.
To find the time at which the ball reaches the ground, we need to set h(t) = 0 and solve for t.
0 = -0.2t + 2t
0 = 1.8t
t = 0
Therefore, the ball reaches the ground after 0 seconds, which means it was kicked from a height of 2 feet.
Suppose h(t) = -0.2t + 2t models the height, in feet, of a ball that is kicked into the air where t is
given as time, in seconds.
After how many seconds does the ball reach its maximum height?
seconds
What is the maximum height of the ball?
feet
seconds
After how many seconds does the ball reach the ground?
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3 answers
clearly, we need a quadratic function, so let's use h(t) = -0.2t^2 + 2t
max height of 5 ft is achieved at t = 5 s
the ball hits the ground at t = 10 s
max height of 5 ft is achieved at t = 5 s
the ball hits the ground at t = 10 s
I apologize for the oversight in the initial calculation. Thank you for providing the corrected values.
Using the quadratic function h(t) = -0.2t^2 + 2t:
To find the time at which the ball reaches its maximum height:
The formula for finding the time at the vertex of a quadratic function in the form h(t) = at^2 + bt + c is t = -b / 2a. In this case, a = -0.2 and b = 2. Plugging in these values, we get t = -2 / 2(-0.2) = 5 seconds. Therefore, the ball reaches its maximum height after 5 seconds.
To find the maximum height of the ball:
Substitute t = 5 into the function h(t) = -0.2t^2 + 2t:
h(5) = -0.2(5)^2 + 2(5) = -0.2(25) + 10 = -5 + 10 = 5 feet. Therefore, the maximum height of the ball is 5 feet.
To find the time at which the ball reaches the ground:
Set h(t) = 0 and solve for t:
0 = -0.2t^2 + 2t
0 = t(-0.2t + 2)
t = 0 or -0.2t + 2 = 0
t = 0 or t = 10
Therefore, the ball reaches the ground after 10 seconds.
Thank you for pointing out the correct values.
Using the quadratic function h(t) = -0.2t^2 + 2t:
To find the time at which the ball reaches its maximum height:
The formula for finding the time at the vertex of a quadratic function in the form h(t) = at^2 + bt + c is t = -b / 2a. In this case, a = -0.2 and b = 2. Plugging in these values, we get t = -2 / 2(-0.2) = 5 seconds. Therefore, the ball reaches its maximum height after 5 seconds.
To find the maximum height of the ball:
Substitute t = 5 into the function h(t) = -0.2t^2 + 2t:
h(5) = -0.2(5)^2 + 2(5) = -0.2(25) + 10 = -5 + 10 = 5 feet. Therefore, the maximum height of the ball is 5 feet.
To find the time at which the ball reaches the ground:
Set h(t) = 0 and solve for t:
0 = -0.2t^2 + 2t
0 = t(-0.2t + 2)
t = 0 or -0.2t + 2 = 0
t = 0 or t = 10
Therefore, the ball reaches the ground after 10 seconds.
Thank you for pointing out the correct values.