If a cannonball is shot directly upward with a velocity of , its height, in feet, above the ground after t seconds is given by . Find the velocity
and the acceleration after t seconds. When does the cannonball reach its maximum height? What is the maximum height the cannonball reaches? When does it hit
the ground?
The velocity after t seconds is
The acceleration after t seconds is a
(Type an integer or a decimal.)
The cannonball reaches its maximum height at .
(Type an integer or a decimal.)
The cannonball reaches a maximum height of .
(Type an integer or a decimal.)
The height of the cannonball when it hits the ground is .
(Type an integer or a decimal.)
The cannonball hits the ground after sec.
(Type an integer or a decimal.)

Ask by Herrera Powers. in the United States

Mar 27,2025

Question

If a cannonball is shot directly upward with a velocity of , its height, in feet, above the ground after t seconds is given by . Find the velocity
and the acceleration after t seconds. When does the cannonball reach its maximum height? What is the maximum height the cannonball reaches? When does it hit
the ground?
The velocity after t seconds is
The acceleration after t seconds is a
(Type an integer or a decimal.)
The cannonball reaches its maximum height at .
(Type an integer or a decimal.)
The cannonball reaches a maximum height of .
(Type an integer or a decimal.)
The height of the cannonball when it hits the ground is .
(Type an integer or a decimal.)
The cannonball hits the ground after sec.
(Type an integer or a decimal.)

Ask by Herrera Powers. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

The height function is given by  
$$
s(t)=208t-16t^2.
$$

**Step 1. Find the velocity $$ v(t) $$.**  
The velocity is the derivative of the position:
$$
v(t)=\frac{d}{dt}[208t-16t^2]=208-32t.
$$

**Step 2. Find the acceleration $$ a(t) $$.**  
The acceleration is the derivative of the velocity:
$$
a(t)=\frac{d}{dt}[208-32t]=-32.
$$

**Step 3. Determine when the cannonball reaches its maximum height.**  
At maximum height, the velocity is zero:
$$
208-32t=0.
$$
Solving for $$ t $$:
$$
32t=208 \quad\Longrightarrow\quad t=\frac{208}{32}=6.5 \text{ sec}.
$$

**Step 4. Find the maximum height.**  
Substitute $$ t=6.5 $$ into the height function:
$$
s(6.5)=208(6.5)-16(6.5)^2.
$$
Calculate:
$$
208(6.5)=1352, \quad (6.5)^2=42.25, \quad 16(42.25)=676.
$$
Thus,
$$
s(6.5)=1352-676=676 \text{ ft}.
$$

**Step 5. Determine when the cannonball hits the ground.**  
The cannonball hits the ground when $$ s(t)=0 $$:
$$
208t-16t^2=0.
$$
Factor out $$ t $$:
$$
t(208-16t)=0.
$$
So, $$ t=0 $$ (launch time) or
$$
208-16t=0 \quad\Longrightarrow\quad 16t=208 \quad\Longrightarrow\quad t=13 \text{ sec}.
$$

**Step 6. The height when it hits the ground.**  
When the cannonball hits the ground, $$ s(t)=0 $$ ft.

**Summary of Answers:**

- The velocity after $$ t $$ seconds is  
  $$
  v(t)=208-32t.
  $$

- The acceleration after $$ t $$ seconds is  
  $$
  a(t)=-32.
  $$

- The cannonball reaches its maximum height at  
  $$
  t=6.5 \text{ sec}.
  $$

- The cannonball reaches a maximum height of  
  $$
  676 \text{ ft}.
  $$

- The height of the cannonball when it hits the ground is  
  $$
  0 \text{ ft}.
  $$

- The cannonball hits the ground after  
  $$
  t=13 \text{ sec}.
  $$
- Velocity after $$ t $$ seconds: $$ v(t) = 208 - 32t $$ - Acceleration after $$ t $$ seconds: $$ a(t) = -32 $$ - Reaches maximum height at $$ t = 6.5 $$ seconds - Maximum height: 676 feet - Hits the ground at $$ t = 13 $$ seconds

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