Suppose you throw a ball straight up from the ground with a velocity of 224 feet per second. as the ball moves up, gravity slows
1 answer:
Answer:
a) 640 ft
b) 6 seconds (the first time)
c) 14 seconds
Step-by-step explanation:
These problems are solved by using the given equation for the height of the thrown ball with the given values, and solving for the unknown value.
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<h3>a. height after a time</h3>To find the height at a specific time, substitute the time value for t in the height formula and do the arithmetic.
For t = 4, we have ...
h = -16t² +224t = (-16(4) +224)(4) = 160(4) = 640
The height after 4 seconds is 640 ft.
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<h3>b. time for a height</h3>The time required to reach a specific height is found by solving the quadratic equation with h equal to the height value. It can be easier if the equation is in vertex form.
h = -16t² +224t = -16(t² -14t) = -16(t² -14t +49) +784 = -16(t -7)² +784
We want the time to a height of 768 ft, so ...
768 = -16(t -7)² +784
-16/-16 = (t -7)² . . . . . . . . subtract 784, divide by -16
±√1 = t -7 . . . . . . . . . take the square root
t = 7 ±1 = 6 or 8 . . . . . . add 7
The ball will be at a height of 768 ft after 6 seconds, and again after 8 seconds.
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<h3>c. time to reach the ground</h3>The equation from part b applies.
0 = -16(t -7)² +784 . . . . . . . . the height is 0 when the ball is on the ground
-784/-16 = (t -7)² = 49 . . . . . subtract 784, divide by -16
√49 = t -7 . . . . . . . . . . . . . take the positive square root
t = 14 . . . . . . . . . . . . . add 7
The ball will return to the ground after 14 seconds.
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