Question 7 (5 points)
A model rocket is launched with an initial speed of . After seconds, its
height, metres, is modelled by the equation
.
a) State the time it takes for the rocket to hit the ground to the nearest teffth. (write
your answer in blank #1)
b) Will the rocket ever reach a height of 60 m ? (write you answer in blank #2)
c) State the height of the rocket after 4 seconds to the nearest tenth. (write your
answer in blank #3)
d) State the maximum height of the rocket to the nearest tenth. (write your answer in
blank #4)
e) State the height of the rocket at launch. (write your answer in blank #5)

Ask by Deleon Moran. in Canada

Mar 26,2025

Question

Question 7 (5 points)
A model rocket is launched with an initial speed of . After seconds, its
height, metres, is modelled by the equation
.
a) State the time it takes for the rocket to hit the ground to the nearest teffth. (write
your answer in blank #1)
b) Will the rocket ever reach a height of 60 m ? (write you answer in blank #2)
c) State the height of the rocket after 4 seconds to the nearest tenth. (write your
answer in blank #3)
d) State the maximum height of the rocket to the nearest tenth. (write your answer in
blank #4)
e) State the height of the rocket at launch. (write your answer in blank #5)

Ask by Deleon Moran. in Canada

Mar 26,2025

UpStudy AI · Accepted Answer

**a)** To find the time when the rocket hits the ground, set  
$$
h(t)=0.5+32t-4.9t^2=0.
$$  
Rearrange the equation:  
$$
-4.9t^2+32t+0.5=0.
$$  
Multiply through by $$-1$$:  
$$
4.9t^2-32t-0.5=0.
$$  
Apply the quadratic formula  
$$
t=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
$$  
with $$a=4.9$$, $$b=-32$$, and $$c=-0.5$$. Compute the discriminant:  
$$
\Delta=(-32)^2-4(4.9)(-0.5)=1024+9.8=1033.8.
$$  
Thus,  
$$
t=\frac{32\pm\sqrt{1033.8}}{9.8}.
$$  
The negative root is discarded leaving  
$$
t\approx\frac{32+\sqrt{1033.8}}{9.8}\approx \frac{32+32.17}{9.8}\approx\frac{64.17}{9.8}\approx6.6.
$$  
_Blank #1 answer:_ $$6.6$$ seconds.

---

**b)** To determine if the rocket ever reaches a height of $$60\,\text{m}$$, we set  
$$
h(t)=60.
$$  
That is,  
$$
0.5+32t-4.9t^2=60.
$$  
Rearrange into quadratic form:  
$$
-4.9t^2+32t-59.5=0.
$$  
Multiply by $$-1$$:  
$$
4.9t^2-32t+59.5=0.
$$  
Calculate the discriminant:  
$$
\Delta=(-32)^2-4(4.9)(59.5)=1024-4(4.9\times59.5).
$$  
First, compute $$4.9\times59.5$$:  
$$
4.9\times59.5\approx291.55.
$$  
Then,  
$$
\Delta=1024-4(291.55)=1024-1166.2\approx-142.2.
$$  
Since the discriminant is negative, there is no real solution; hence, the rocket never reaches $$60\,\text{m}$$.  
_Blank #2 answer:_ No.

---

**c)** The height at $$4$$ seconds is found by evaluating  
$$
h(4)=0.5+32(4)-4.9(4)^2.
$$  
Compute step by step:  
$$
32(4)=128,\quad (4)^2=16,\quad 4.9(16)=78.4.
$$  
Thus,  
$$
h(4)=0.5+128-78.4=50.1.
$$  
Rounded to the nearest tenth,  
_Blank #3 answer:_ $$50.1\,\text{m}$$.

---

**d)** The maximum height occurs at the vertex of the parabola. For a quadratic $$h(t)=-4.9t^2+32t+0.5$$, the time coordinate of the vertex is  
$$
t=-\frac{b}{2a}=-\frac{32}{2(-4.9)}=\frac{32}{9.8}\approx3.3\,\text{seconds}.
$$
Now, evaluate $$h(3.3)$$:  
$$
h(3.3)=0.5+32(3.3)-4.9(3.3)^2.
$$  
Compute:  
$$
32(3.3)=105.6,\quad (3.3)^2\approx10.89,\quad 4.9(10.89)\approx53.4.
$$
Therefore,  
$$
h(3.3)=0.5+105.6-53.4\approx52.7.
$$
Rounded to the nearest tenth,  
_Blank #4 answer:_ $$52.7\,\text{m}$$.

---

**e)** The height of the rocket at launch is given by evaluating the model at $$t=0$$:  
$$
h(0)=0.5+32(0)-4.9(0)^2=0.5.
$$
Thus,  
_Blank #5 answer:_ $$0.5\,\text{m}$$.
**a)** The rocket hits the ground after approximately **6.6 seconds**. **b)** No, the rocket never reaches a height of **60 meters**. **c)** After 4 seconds, the rocket is at a height of **50.1 meters**. **d)** The maximum height reached by the rocket is **52.7 meters**. **e)** At launch, the rocket is at a height of **0.5 meters**.

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