Question

1. a nerf gun shoots a foam dart into the air. the following function models the height, ( f(x) ), in feet, of an object ( x ) seconds after it is in the air: ( f(x)=-16 x^{2}+16 x+6 ).

a. is the foam dart going up or down after 1 second?
b. how high is the foam dart at 2 seconds? can this happen?

1. a ball is thrown into the air from a height of 4 feet at time ( t = 0 ). the function that models this situation is ( h(t)=-16 t^{2}+63 t + 4 ), where ( t ) is measured in seconds and ( h ) is the height in feet.

a. what is the height of the ball after 2 seconds?
b. when will the ball reach a height of 50 feet?
c. what is the maximum height of the ball?
d. when will the ball hit the ground?
e. what domain makes sense for the function?
the function ( s(t)=v t+h - 0.5 a t^{2} ) represents the height of an object, ( s ), from the ground after time, ( t ), when the object is thrown with an initial velocity of ( v ), at an initial height of ( h ), and where ( a ) is the acceleration due to gravity (32 feet per second squared). a baseball player hits a baseball 4 feet above the ground with an initial velocity of 80 feet per second. about how long will it take the baseball to hit the ground?
a. 2 seconds
b. 3 seconds
c. 4 seconds
d. 5 seconds

1. a flying disk is thrown into the air from a height of 25 feet at time ( t = 0 ). the function that models this situation is ( h(t)=-16 t^{2}+75 t + 25 ), where ( t ) is measured in seconds and ( h ) is the height in feet. what values of ( t ) best describe the times when the disk is flying in the air?

a. ( 0<t<5 )
b. ( 0<t<25 )
c. all real numbers
d. all positive integers

1. an object is thrown in the air with an initial velocity of ( 5 mathrm{~m} / mathrm{s} ) from a height of ( 9 mathrm{~m} ). the equation ( h(t)=-4.9 t^{2}+5 t + 9 ) models the height of the object in meters after ( t ) seconds. how many seconds does it take for the object to hit the ground?

a. ( 0.94 ) seconds
b. ( 1.77 ) seconds
c. ( 1.96 ) seconds
d. ( 9.0 ) seconds

Explanation:

Step1: Substitute \(t = 0\) into \(s(t)\)

The function \(s(t)=vt + h-0.5at^{2}\). Given \(v = 80\) (initial velocity), \(h = 4\) (initial height), \(a = 32\) (acceleration due to gravity). When the ball hits the ground, \(s(t)=0\). So the equation becomes \(0 = 80t+4 - 0.5\times32t^{2}\).
Simplify the equation: \(0 = 80t + 4-16t^{2}\), or \(16t^{2}-80t - 4=0\). Divide through by \(4\) to get \(4t^{2}-20t - 1=0\).

Step2: Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

For the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 4\), \(b=-20\), \(c = - 1\)).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-20)^{2}-4\times4\times(-1)=400 + 16=416\).
Then \(t=\frac{20\pm\sqrt{416}}{8}=\frac{20\pm4\sqrt{26}}{8}=\frac{5\pm\sqrt{26}}{2}\).
We take the positive root \(t=\frac{5+\sqrt{26}}{2}\approx\frac{5 + 5.1}{2}=\frac{10.1}{2}=5.05\) (approximate value). But let's check another way.
If we use the function \(h(t)=-16t^{2}+v_{0}t + h_{0}\) (where \(v_{0}=80\), \(h_{0}=4\)).
Set \(h(t)=0\), so \(-16t^{2}+80t + 4=0\), \(4t^{2}-20t - 1 = 0\).
Using the quadratic formula \(t=\frac{20\pm\sqrt{400+16}}{8}=\frac{20\pm\sqrt{416}}{8}\approx\frac{20 + 20.4}{8}\approx5\) (approximate by estimating \(\sqrt{416}\approx20.4\)).

For problem 8:

Step1: Set \(h(t)=0\)

Given \(h(t)=-16t^{2}+75t + 25\). Set \(h(t)=0\), so \(-16t^{2}+75t + 25=0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=-16\), \(b = 75\), \(c = 25\).
\(\Delta=b^{2}-4ac=(75)^{2}-4\times(-16)\times25=5625+1600 = 7225\).
\(t=\frac{-75\pm\sqrt{7225}}{-32}=\frac{-75\pm85}{-32}\).
We get two roots: \(t_{1}=\frac{-75 + 85}{-32}=\frac{10}{-32}\) (rejected since \(t>0\)) and \(t_{2}=\frac{-75-85}{-32}=\frac{-160}{-32}=5\).
The disk is in the air when \(h(t)>0\), so \(0 < t<5\).

For problem 9:

Step1: Set \(h(t)=0\)

Given \(h(t)=-4.9t^{2}+5t + 9\). Set \(h(t)=0\), so \(-4.9t^{2}+5t + 9=0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=-4.9\), \(b = 5\), \(c = 9\).
\(\Delta=b^{2}-4ac=(5)^{2}-4\times(-4.9)\times9=25 + 176.4=201.4\).
\(t=\frac{-5\pm\sqrt{201.4}}{-9.8}\). We take the positive root \(t=\frac{-5+\sqrt{201.4}}{-9.8}\approx\frac{-5+14.2}{-9.8}=\frac{9.2}{-9.8}\) (wrong sign). Correct formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-5\pm\sqrt{25 + 176.4}}{-9.8}=\frac{-5\pm\sqrt{201.4}}{-9.8}\). Taking the positive root \(t=\frac{-5+\sqrt{201.4}}{-9.8}\approx\frac{-5 + 14.2}{-9.8}\) (error in sign, actually \(t=\frac{-5\pm\sqrt{25+176.4}}{2\times(-4.9)}=\frac{-5\pm\sqrt{201.4}}{-9.8}\). The positive root \(t=\frac{5+\sqrt{201.4}}{9.8}\approx\frac{5+14.2}{9.8}=\frac{19.2}{9.8}\approx1.96\).

Answer:

For the baseball problem (similar to problem 7): The answer is D. 5 seconds (approximate value).
For problem 8: A. \(0 < t<5\)
For problem 9: C. 1.96 seconds