The height of a ball thrown vertically upward from a rooftop is modelled by y=-5t^2+20t+50 , where h(t) is the ball's height above the ground , in meters , at time t seconds after the throw
a) Determine the max height of the ball
b) how long does it take for the ball to reach its max height
c) How high is the rooftop
y = 5(10 + 4t - t^2)
max height achieved where t = -b/2a = 2
y(2) = -5(4) + 20(2) + 50 = 70
as above, when t=2
The rooftop is the ball's initial height. That is, when t=0. So, 50 ft.
a) Well, to determine the maximum height of the ball, we need to find the vertex of the quadratic equation. Remember, the vertex is given by the formula (-b/2a, f(-b/2a)). In this case, a = -5 and b = 20. Plugging these values in, we get (-20/2(-5), f(-20/2(-5))). Simplifying, we have (2, f(2)). To find f(2), we substitute t = 2 into the equation y = -5t^2 + 20t + 50, giving us y = -5(2)^2 + 20(2) + 50. Do the math, and you'll find the maximum height of the ball.
b) The time it takes for the ball to reach its maximum height is the x-coordinate of the vertex. So, in this case, it's just -b/2a. Plug in the values, and voila, you have your answer!
c) To determine the height of the rooftop, we need to find the value of y when t = 0 (since we're interested in the height from the ground). Plug in t = 0 into the equation y = -5t^2 + 20t + 50, and you'll find the height of the rooftop.
a) To determine the maximum height of the ball, we need to find the vertex of the parabolic equation. The vertex is given by the equation:
t = -b/2a
where a, b, and c are the coefficients of the quadratic equation. In this case, a = -5, b = 20, and c = 50. Plugging these values into the equation, we get:
t = -20 / (2 * -5)
t = 2 seconds
Now we can find the height at this time by substituting t = 2 into the equation for y:
y = -5(2)^2 + 20(2) + 50
y = -5(4) + 40 + 50
y = -20 + 40 + 50
y = 70 meters
Therefore, the maximum height of the ball is 70 meters.
b) We have already found that the time it takes for the ball to reach its maximum height is 2 seconds.
c) The height of the rooftop can be found by substituting t = 0 into the equation for y. This is because at the start of the motion (t = 0), the ball is at rooftop level. Substituting t = 0 into the equation, we get:
y = -5(0)^2 + 20(0) + 50
y = 50 meters
Therefore, the height of the rooftop is 50 meters.
To determine the maximum height of the ball, we need to find the vertex of the quadratic function y = -5t^2 + 20t + 50. The vertex is the highest point on the parabolic path of the ball.
a) To find the maximum height, we can use the formula t = -b/2a to find the time at which the ball reaches its highest point. In this case, a = -5 and b = 20:
t = -20 / (2 * -5)
t = -20 / -10
t = 2
Now, substitute this value of t back into the equation to find the corresponding maximum height:
y = -5(2)^2 + 20(2) + 50
y = -5(4) + 40 + 50
y = -20 + 40 + 50
y = 70
So, the maximum height of the ball is 70 meters.
b) The time it takes for the ball to reach its maximum height is 2 seconds.
c) To determine the height of the rooftop, we need to find the initial height of the ball when t = 0. In the equation y = -5t^2 + 20t + 50, substitute t = 0:
y = -5(0)^2 + 20(0) + 50
y = 0 + 0 + 50
y = 50
Therefore, the height of the rooftop is 50 meters.