a science class designed a ball launcher and tested it by shooting a tennis ball straight up from the top of a 15 story building. they determined that 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 of the ball above the ground at time t what is the maximum height of the ball?at what time will the ball hit the ground? over what domain is the ball rising and over what domain is the ball falling with graphing shown
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To find the maximum height of the ball, we need to find the vertex of the quadratic function h(t) = -16t^2 + 144t + 160. The vertex of a quadratic function in the form ax^2 + bx + c is at the point (-b/2a, f(-b/2a)), so for our function:
t = -144 / 2(-16) = -144 / -32 = 4.5
To find the maximum height, we can substitute t = 4.5 into the function h(t):
h(4.5) = -16(4.5)^2 + 144(4.5) + 160 = -324 + 648 + 160 = 484 feet
Therefore, the maximum height of the ball is 484 feet.
To find when the ball will hit the ground, we need to find when the height is 0. So we set h(t) = -16t^2 + 144t + 160 equal to 0:
-16t^2 + 144t + 160 = 0
We can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives:
-16(t^2 - 9t - 10) = 0
-16(t - 10)(t + 1) = 0
This gives us t = 10 or t = -1. Since time cannot be negative, the ball will hit the ground at t = 10 seconds.
The ball is rising for the time interval [0, 4.5] and falling for the time interval [4.5, 10].