A pomegranate is thrown from a ground level straight up into the air at time t = 0 with velocity 67 feet per second. Its height at time t seconds is f(t)=-16^2+67t.
(a) Find the time it hits the ground and (b) the time it reaches its highest point. (c) What is the maximum height?
Please round to two decimal places.
a) when it hits the ground, f(t) = 0
0 = -16t^2 + 67t
16t^2 - 67t = 0
t(16t - 67) = 0
t = 0 or t = 67/16 = appr 4.19 seconds
t=0 would be the start of the toss
so the answer you want is t = 4.19 sec
b) since you labeled it "pre-calculus" I assume you cannot at this point take the derivative, too bad
so we have to complete the square
f(t) = -16t^2 + 67t
= -16(t^2 - (67/16)t + 4489/1024 - 4489/1024)
= -16( (t - 67/32)^2 - 4489/1024)
= -16(t - 67/32)^2 + 4489/64
it will reach the maximum height at 67/32 or appr 2.09 sec, and that max height is 4489/64 or 70.14 ft
Notice we could have used the properties of the parabola to shorten this up a bit
since the x-intercepts are 0 and 4.19, the vertex must lie midway between them
This would be 2.09 , the answer I got when completing the square. Subbing 2.09 in for f(2.09) gives us
-16(2.09)^2 + 67(2.09) = 70.14
To find the time when the pomegranate hits the ground, we need to find the time when its height is equal to zero. To do that, we set the equation f(t) = -16t^2 + 67t equal to zero and solve for t.
-16t^2 + 67t = 0
Factoring out t, we get:
t(-16t + 67) = 0
Setting each factor equal to zero, we get:
t = 0 or -16t + 67 = 0
The first solution, t = 0, corresponds to the initial time when the pomegranate was thrown, so it is not the time it hits the ground. Therefore, we focus on the second equation:
-16t + 67 = 0
Adding 16t to both sides, we have:
67 = 16t
Dividing both sides by 16, we get:
t = 67/16
So, the time it hits the ground is approximately 4.19 seconds (rounded to two decimal places).
To find the time when the pomegranate reaches its highest point, we need to find the vertex of the parabolic equation f(t) = -16t^2 + 67t. The vertex of a parabola with equation f(t) = at^2 + bt + c is given by the coordinates (-b/2a, f(-b/2a)). In this case, a = -16 and b = 67.
Using the formula, we find:
t = -67 / (2 * -16)
Simplifying, we get:
t = 67/32
So, the time it reaches its highest point is approximately 2.09 seconds (rounded to two decimal places).
To find the maximum height, we substitute the time t = 67/32 into the equation f(t) = -16t^2 + 67t.
f(67/32) = -16(67/32)^2 + 67(67/32)
Simplifying, we get:
f(67/32) = -16(4489/1024) + (67*67)/32
f(67/32) = -70592/1024 + 4489/32
f(67/32) ≈ -69.051 + 140.281
f(67/32) ≈ 71.23
So, the maximum height is approximately 71.23 feet (rounded to two decimal places).