A ball is shot straight up from the top of a 15-story building. The motion of the ball could be described by the function: h(t)= -16t^2 + 144t + 160 where t represents the time the ball is in the air in seconds and h(t) represents the height in feet .
What is the height of the ball after 4 seconds?
What is the maximum height of the ball?
At what time will the ball hit the ground?
What is the height of the ball after 4 seconds?
Just plug in t=4
What is the maximum height of the ball?
recall that for ax^2+bx+c the vertex is at x = -b/2a
Also, less well known is that the y-coordinate of the vertex is (4ac-b^2)/4a
At what time will the ball hit the ground?
just solve for t when h=0
To find the height of the ball after 4 seconds, substitute t = 4 into the function h(t):
h(4) = -16(4)^2 + 144(4) + 160
= -16(16) + 576 + 160
= -256 + 576 + 160
= 480 feet
Therefore, the height of the ball after 4 seconds is 480 feet.
To find the maximum height of the ball, we need to find the vertex of the quadratic function represented by h(t). The vertex of a quadratic function in the form ax^2 + bx + c is given by the equation:
t = -b / (2a)
For h(t) = -16t^2 + 144t + 160, a = -16 and b = 144. Substituting these values into the formula:
t = -144 / (2*(-16))
= -144 / (-32)
= 4.5
So the maximum height of the ball occurs at t = 4.5 seconds.
To find the time when the ball hits the ground, we need to find the value of t when h(t) = 0. So we can set the equation h(t) = 0 and solve for t:
-16t^2 + 144t + 160 = 0
Factoring this equation, we get:
-16(t^2 - 9t - 10) = 0
Setting each factor equal to zero, we have:
t^2 - 9t - 10 = 0
Now we can solve this quadratic equation using factoring or the quadratic formula. Factoring it, we have:
(t - 10)(t + 1) = 0
Setting each factor equal to zero, we have:
t - 10 = 0 or t + 1 = 0
Solving each equation, we get:
t = 10 or t = -1
Since time cannot be negative in this context, we reject t = -1. Therefore, the ball hits the ground at t = 10 seconds.
To find the height of the ball after 4 seconds, we plug in t = 4 into the function h(t) = -16t^2 + 144t + 160 and calculate the result.
h(4) = -16(4)^2 + 144(4) + 160
= -16(16) + 576 + 160
= -256 + 576 + 160
= 480 feet
Therefore, the height of the ball after 4 seconds is 480 feet.
To find the maximum height of the ball, we need to determine the vertex of the parabolic function h(t) = -16t^2 + 144t + 160. The vertex of a parabola with equation f(x) = ax^2 + bx + c is located at the point (-b/2a, f(-b/2a)).
In this case, a = -16, b = 144, and c = 160. Therefore, the time value of the vertex is -144/(2*(-16)) = -144/-32 = 4.5 seconds. Plugging this value into the function, we can find the height.
h(4.5) = -16(4.5)^2 + 144(4.5) + 160
= -16(20.25) + 648 + 160
= -324 + 648 + 160
= 484 feet
Hence, the maximum height of the ball is 484 feet.
To determine when the ball hits the ground, the height h(t) must equal zero. We can set the height function to zero and solve for t.
-16t^2 + 144t + 160 = 0
We can either factor this equation or use the quadratic formula. Factoring is generally not straightforward in this case, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = -16, b = 144, and c = 160, we get:
t = (-144 ± √(144^2 - 4(-16)(160))) / (2(-16))
= (-144 ± √(20736 + 10240)) / (-32)
= (-144 ± √(30976)) / (-32)
= (-144 ± 176) / (-32)
Simplifying this further, we have two solutions:
t1 = (-144 + 176) / (-32) = 1
t2 = (-144 - 176) / (-32) = 9
Since time cannot be negative in this context, the ball hits the ground at t = 9 seconds.
Therefore, the ball hits the ground 9 seconds after being shot up, at time t = 9 seconds.