A softball player has hit a ball into the air. The height, y, of the ball x seconds after it is hit is given by y = -16x^2 + 78x +3.25
-What is the initial height of the softball?
-How long was the softball in the air?
-What is the maximum height of the softball?
-How long did it take for the softball to reach its maximum height?

1) 3.25 units

2) 4.875 seconds

3) 98.3125 units

4) 2.4375 seconds

Step-by-step explanation:

1) The given equation of motion of the soft ball is y = -16·x² + 78·x + 3.25

Where;

y = The height of the softball

x = The time in seconds after it is hit

Therefore, the initial height of the softball is given by the value of y when x = 0 as follows;

y₀ = -16 × 0² + 78 × 0 + 3.25 = 3.25

Therefore, the initial height of the softball is 3.25 units

2) Given that the initial height of the soft ball is 3.25 ft., we have;

y = -16·x² + 78·x + 3.25 = x·(-16·x + 78) + 3.25

Therefore, the times at which the level of the softball is at 3.25 unit height, are;

x = 0 and ((-16·x + 78) = 0 which gives x = -78/(-16) = 4.875) 4.875 seconds

Therefore, from start time x = 0, the softball was in the air for 4.875 seconds

3) Given that the equation of motion of the softball is the equation of a parabola, the maximum height of the ball is given by the x-coordinate value at the top of the curve where the slope is zero as follows;

d(-16·x² + 78·x + 3.25)/dx = -32·x + 78 = 0

x = -78/(-32) = 39/16 = 2⁷/₁₆ = 2.4375 seconds

The maximum height of the softball is therefore;

 = -16 × 2.4375² + 78 × 2.4375 + 3.25 = 98.3125

The maximum height of the softball is 98.3125 units

4) From the calculation for the maximum height, the time it took the softball to reach maximum height = 2.4375 seconds

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answer

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step-by-step explanation: