A: To find out how long the ball was in the air, we need to find the time when the height of the ball is 0. This means setting h=0 in the equation:
0 = -8t^2 + 18t + 5
This is a quadratic equation that can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values a=-8, b=18, and c=5:
t = (-18 ± √(18^2 - 4(-8)(5))) / 2(-8)
t = (-18 ± √(324 + 160)) / -16
t = (-18 ± √484) / -16
t = (-18 ± 22) / -16
This gives two possible values for t: t=1/2 and t=5/2. Since time cannot be negative, the ball was in the air for 1/2 seconds.
B: To find how high the ball got, we plug in the time the ball was in the air into the equation for height:
h = -8(1/2)^2 + 18(1/2) + 5
h = -8(1/4) + 9 + 5
h = -2 + 9 + 5
h = 12 feet
Therefore, the soccer ball reached a height of 12 feet.
C: The highest point of the ball's trajectory occurs at the vertex of the parabola given by the equation. The x-coordinate of the vertex is given by -b/2a:
t = -18 / 2(-8)
t = 18 / 16
t = 9/8
Now we need to find the height at this time:
h = -8(9/8)^2 + 18(9/8) + 5
This simplifies to:
h = -8(81/64) + 81/4 + 5
h = -162/16 + 324/16 + 80/16
h = 242/16
h = 15.125 feet
Therefore, the ball hit its highest point at t = 9/8 seconds, reaching a height of 15.125 feet.
Your friend tosses a ball into the air at an initial velocity of 18 feet per second. The equation h= -8t^2+18t+5 models the height h of the ball t seconds after it was thrown.
A: How long was the ball in the air?
B: How high did the soccer ball get?
C: When did the ball hit its highest point?
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