The height of a ball above the ground in feet is defined by the function \( h(t) = -16t^2 + 80t + 3 \), where \( t \) is the number of seconds after the ball is thrown.
What is the value of \( h(t) \) two seconds after the ball is thrown?
Answer (3)
plug in 2 for t. => -16*(2^2) +80 2 +3 => -16 4 +160 +3 => -64+163 => 99
"The value of h(t), two seconds after the ball is thrown, is 131 feet.
To find the height of the ball two seconds after it is thrown, we substitute t = 2 into the height function h ( t ) = − 16 t 2 + 80 t + 3 .[/tex]
First, we calculate t^2 \) when \( t = 2 \
t 2 = 2 2 = 4
Next, we substitute [tex] t = 2 and t 2 = 4 into the function:
h ( 2 ) = − 16 ( 4 ) + 80 ( 2 ) + 3
Now, we perform the multiplication:
h ( 2 ) = − 64 + 160 + 3
Then, we add the numbers together:
h ( 2 ) = 96 + 3
Finally, we sum the constants to find the height:
h ( 2 ) = 131
Therefore, two seconds after the ball is thrown, its height above the ground is 131 feet."
The height of the ball two seconds after it is thrown, calculated using the function, is 99 feet. This was done by substituting t = 2 into the height equation and performing the arithmetic. The final result shows that the height at that time is 99 feet.
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