The height above the ground of a ball thrown up with a velocity of 96 feet per second from a height of 6 feet is 6+96t-16^2 feet, where t is the time in seconds. According to this model, how high is the ball after 7 seconds? Explain (Thank you in advance! :)
set
6 = 6 + 96t - 16t^2
16t^2 - 96t = 0
16t(t - 6t)=0
t = 0 or t = 6
the ball was 6 feet high at the beginning (t=0) and
on it way back down at 6 seconds later.
(I bet the ball was at its highest point after 3 seconds)
To find the height of the ball after 7 seconds, we'll substitute t = 7 into the given equation:
Height = 6 + 96t - 16t^2
Now, let's plug in t = 7:
Height = 6 + 96*7 - 16*(7^2)
Simplifying the equation:
Height = 6 + 672 - 16*49
= 6 + 672 - 784
= -106
According to this model, the height of the ball after 7 seconds is -106 feet.
However, it's important to note that a negative value for height doesn't make sense in this case. It suggests that the ball is below ground level, which is not physically possible in this scenario. Therefore, there may be an error in the model or the equation provided.
To find out how high the ball is after 7 seconds, we need to substitute the value of "t" into the equation and simplify.
Given: the equation for the height of the ball is h = 6 + 96t - 16t^2
We substitute t = 7 into the equation:
h = 6 + 96(7) - 16(7)^2
Now, let's solve this equation step by step.
First, we calculate 96(7) which equals 672.
h = 6 + 672 - 16(7)^2
Next, we calculate 7 raised to the power of 2 (7^2), which equals 49.
h = 6 + 672 - 16(49)
Then, we calculate 16 times 49, which equals 784.
h = 6 + 672 - 784
Now, we add 6 and 672:
h = 678 - 784
Finally, subtracting 784 from 678 gives us:
h = -106
Therefore, according to this model, the height above the ground of the ball after 7 seconds is -106 feet.