A ball is tossed upward from the ground. Its height

Rivka Thorpe 2021-08-11 Answered
A ball is tossed upward from the ground. Its height in feet above ground after t seconds is given by the function h(t)=16t2+24t. Find the maximum height of the ball and the number of seconds it took for the ball to reach the maximum height.
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Expert Answer

Viktor Wiley
Answered 2021-08-12 Author has 84 answers
1) Concept:
The maximum height of the ball is the maximum value of h(t). The degree of the function h(t)=16t2+24t is 2 and hence it is a quadratic function. The graphical representation of the quadratic function is a parabola. The coefficient of the t2 is -16. Therefore, it is a parabola that opens downward. So, the h(t) or the y coordinate of the vertex gives the maximum height and the x coordinate of the vertex gives the number of seconds it took for the ball to reach the maximum height.
Step 1: Compare h(t) with the general quadratic equation ax2+bx+c. Find the values of a and b.
Step 2: Find the vertex using the formula (b2a,h(b2a)).
Step 3: The x coordinate of the vertex is the numbers of seconds the ball takes to reach the maximum height and the y coordinate of the vertex is the maximum height of the ball.
2) Calculation:
Comparing h(t)=16t2+24t with ax2+bx+c, we have a=16.b=24.
Now, to find the x-coordinate of the vertex, substitute the values of a and b in b2a.
b2a=242(16)
24 and 16 are multiples of 2. So,we can write them as factors of 2 and the common factors can be cancelled out. Since both the numerator and the denominator has negative sign, the fraction becomes positive.
242(16)=2×2×2×32×2×2×2×2=34
Now, find h(34). Substitute 34 for t in the function h(t)=16t2+24t.
h(34)=16(34)2+24(34)
Again, write the numbers as product of their factors.
h(34)=2×2×2×2×(3×34×4)+2×2×2×3(34)
Cancelling out the common factors in the numerator and the denominator, we have,
h(32)=2×2×2×2(3×32×2×2×2)+2×2×2×3(32×2)
Evaluate using order of operations.
h(32)=3×3+2×3×3
h(32)=9+18=9
Now, we have,
b2a=34 and h(b2a)=9.
Therefore, the vertex is (34,9)
So, it took 0.75 seconds for the ball to reach the ground.
Conclusion:
The maximum height of the ball is 9 feet and it took 34 or 0.75 seconds for the ball to reach the ground.
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