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.

Question

What is the maximum height of the ball?
When will the ball hit the ground?

Answer 1

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.

Answer 2

Are you sure

Answer 3

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.

Answer 4

The height is wrong

Answer 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.