Question
Shawn shoots a basketball but whoops! There goes an airball. The ball's trajectory can be modeled by the function h(t)=-t^2+3t+10 where h is the height in feet and t is the time in seconds.
What is the maximum height of the ball?
When will the ball hit the ground?
What is the maximum height of the ball?
When will the ball hit the ground?
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The height is wrong
Answered by
GPT 3.5
To find the maximum height of the ball, we need to find the vertex of the parabolic function h(t)=-t^2+3t+10.
The vertex of a parabola in the form y=ax^2+bx+c is given by the point (-b/2a, f(-b/2a)). In this case, a=-1, b=3, and c=10.
So, the t-coordinate of the vertex is -b/2a = -3/(2*-1) = 3/2. Substituting t=3/2 into h(t)=-t^2+3t+10 gives us the maximum height:
h(3/2) = -(3/2)^2 + 3(3/2) + 10
h(3/2) = -9/4 + 9/2 + 10
h(3/2) = 11/4
Therefore, the maximum height of the ball is 11/4 feet.
To find when the ball hits the ground, we need to set h(t)=-t^2+3t+10 equal to 0 and solve for t:
0 = -t^2+3t+10
t^2 - 3t - 10 = 0
(t-5)(t+2) = 0
So, t=5 or t=-2. Since time cannot be negative in this context, the ball will hit the ground at t=5 seconds.
The vertex of a parabola in the form y=ax^2+bx+c is given by the point (-b/2a, f(-b/2a)). In this case, a=-1, b=3, and c=10.
So, the t-coordinate of the vertex is -b/2a = -3/(2*-1) = 3/2. Substituting t=3/2 into h(t)=-t^2+3t+10 gives us the maximum height:
h(3/2) = -(3/2)^2 + 3(3/2) + 10
h(3/2) = -9/4 + 9/2 + 10
h(3/2) = 11/4
Therefore, the maximum height of the ball is 11/4 feet.
To find when the ball hits the ground, we need to set h(t)=-t^2+3t+10 equal to 0 and solve for t:
0 = -t^2+3t+10
t^2 - 3t - 10 = 0
(t-5)(t+2) = 0
So, t=5 or t=-2. Since time cannot be negative in this context, the ball will hit the ground at t=5 seconds.
Answered by
GPT 3.5
I apologize for the error in my previous response. Let me correct that:
To find the maximum height of the ball, we need to find the vertex of the parabolic function h(t)=-t^2+3t+10.
The vertex of a parabola in the form y=ax^2+bx+c is given by the point (-b/2a, f(-b/2a)). In this case, a=-1, b=3, and c=10.
So, the t-coordinate of the vertex is -b/2a = -3/(2*-1) = 3/2. Substituting t=3/2 into h(t)=-t^2+3t+10 gives us the maximum height:
h(3/2) = -(3/2)^2 + 3(3/2) + 10
h(3/2) = -9/4 + 9/2 + 10
h(3/2) = 17/4
Therefore, the maximum height of the ball is 17/4 = 4.25 feet.
Now, to find when the ball hits the ground, we need to set h(t)=-t^2+3t+10 equal to 0 and solve for t:
0 = -t^2+3t+10
t^2 - 3t - 10 = 0
(t-5)(t+2) = 0
So, t=5 or t=-2. Since time cannot be negative in this context, the ball will hit the ground at t=5 seconds.
To find the maximum height of the ball, we need to find the vertex of the parabolic function h(t)=-t^2+3t+10.
The vertex of a parabola in the form y=ax^2+bx+c is given by the point (-b/2a, f(-b/2a)). In this case, a=-1, b=3, and c=10.
So, the t-coordinate of the vertex is -b/2a = -3/(2*-1) = 3/2. Substituting t=3/2 into h(t)=-t^2+3t+10 gives us the maximum height:
h(3/2) = -(3/2)^2 + 3(3/2) + 10
h(3/2) = -9/4 + 9/2 + 10
h(3/2) = 17/4
Therefore, the maximum height of the ball is 17/4 = 4.25 feet.
Now, to find when the ball hits the ground, we need to set h(t)=-t^2+3t+10 equal to 0 and solve for t:
0 = -t^2+3t+10
t^2 - 3t - 10 = 0
(t-5)(t+2) = 0
So, t=5 or t=-2. Since time cannot be negative in this context, the ball will hit the ground at t=5 seconds.
Answered by
GPT 3.5
I apologize for the mistake. Let's correct that:
To find the maximum height of the ball, we need to find the vertex of the parabolic function h(t)=-t^2+3t+10.
The t-coordinate of the vertex is given by -b/2a, where a = -1 and b = 3 in this case.
t = -b / (2a) = -3 / (2*(-1)) = 3/2
Substitute t = 3/2 into the function h(t) to find the maximum height:
h(3/2) = -(3/2)^2 + 3(3/2) + 10
h(3/2) = -9/4 + 9/2 + 10
h(3/2) = -9/4 + 18/4 + 40/4
h(3/2) = 49/4
Therefore, the maximum height of the ball is 49/4 = 12.25 feet.
To find when the ball hits the ground, we need to set h(t) = -t^2 + 3t + 10 equal to 0 and solve for t:
0 = -t^2 + 3t + 10
t^2 - 3t - 10 = 0
(t - 5)(t + 2) = 0
t = 5 or t = -2
Since time cannot be negative in this context, the ball will hit the ground at t = 5 seconds.
To find the maximum height of the ball, we need to find the vertex of the parabolic function h(t)=-t^2+3t+10.
The t-coordinate of the vertex is given by -b/2a, where a = -1 and b = 3 in this case.
t = -b / (2a) = -3 / (2*(-1)) = 3/2
Substitute t = 3/2 into the function h(t) to find the maximum height:
h(3/2) = -(3/2)^2 + 3(3/2) + 10
h(3/2) = -9/4 + 9/2 + 10
h(3/2) = -9/4 + 18/4 + 40/4
h(3/2) = 49/4
Therefore, the maximum height of the ball is 49/4 = 12.25 feet.
To find when the ball hits the ground, we need to set h(t) = -t^2 + 3t + 10 equal to 0 and solve for t:
0 = -t^2 + 3t + 10
t^2 - 3t - 10 = 0
(t - 5)(t + 2) = 0
t = 5 or t = -2
Since time cannot be negative in this context, the ball will hit the ground at t = 5 seconds.
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