Question

HELP, THIS IS WAS DUE SO LONG AGO. I PUT THE DIAGRAM THIS TIME.

1.) What is the height of the cannon before it is launched, at t=0? Remember to include units.



2.) A projectile’s maximum height is shown by the vertex of the parabola. When does the cannonball reach its maximum height? Round to the nearest whole number and remember to include units.



3.) What is the maximum height of the cannonball? Round to the nearest whole number and remember to include units.



4.) How long does it take the cannonball to land on the ground? Round your answer to the nearest whole number and remember to include your units.



5.) If you launch the cannonball from the ground instead (hi=0), what does that change in the original equation?



6.) What would be the equation of the cannonball’s path if it was launched with the same velocity, but from the ground instead?



7.) *BONUS* Using this new equation, what is the new maximum height of the ball and how long did it take to hit the ground? Round to the nearest whole number and remember to include units.

Answer 1

1) 58 m

2) t = 2 s

3) 78 m

4) t = 6 s

5) It changes the y-intercept (58) to zero

6)
`h(t)=-4.9t^2+19.8t`

7) h = 20 m

t = 4 s

Explanation:

`h(t)=-4.9t^2+19.8t+58`

1) When t = 0:

`\implies h(0)=-4.9(0^2)+19.8(0)+58=58`

Therefore, the height of the cannon before it is launched is 58 m

2) To find the time when the parabola reaches its max height, differentiate the function with respect to t, equal to zero, then solve for t:

`\implies h'(t)=-9.8t+19.8`

`\implies h'(t)=0 \\\\\implies-9.8t+19.8=0\\\\\implies t=2.020408163...\\\\\implies t=2 \ \textsf{s (nearest whole number)}`

3) The maximum height of the cannonball is when t = 2.020408163...:

`\implies h(2.020408163...)=-4.9(2.020408163...^2)+19.8(2.020408163...)+58=78.0020408...`

Therefore, the maximum height of the cannonball is 78 m (nearest whole number)

4) Set the function to zero and solve for t:

`\implies-4.9t^2+19.8t+58=0\\\\\implies t=-1.96942..., t=6.01024...`

As time is positive, t = 6 s (nearest whole number)

5) If you launch the cannonball from the ground, where
`h_i=0`, then the y-intercept will be 0 and the function will be
`h(t)=-4.9t^2+19.8t`

6)
`h(t)=-4.9t^2+19.8t`

7) To find the maximum height of the parabola, differentiate the function with respect to t, equal to zero, solve for t, then substitute the found value for t into the equation:

`\implies h(t)=-9.8t+19.8`

`\implies h'(t)=0 \\\\\implies-9.8t+19.8=0\\\\\implies t=2.020408163...\\\\\implies t=2 \ \textsf{s (nearest whole number)}`

`\implies h(2.020408163...)=-4.9(2.020408163...)^2+19.8(2.020408163...)=20.00204081...`
`\implies h = 20 \textsf{ m (nearest whole number)}`

To find the time it took to hit the ground, set the function to zero and solve for t:

`\implies -4.9t^2+19.8t=0\\\\\implies t=0, t=4.04081632653...`

`\implies t=4 \textsf{ s (nearest whole number)}`

Answer 2

• h(t)=-4.9t^2+19.8t+58
• Assuming for same unit length

#1

put t=0 in function

`\\ \rm\hookrightarrow h(0)=58m`

#2

It's already located

• Vertex(4,79)

`\\ \rm\hookrightarrow T\:at\:H_(max)=4s`

#3

`\\ \rm\hookrightarrow H_(max)=79m`

#4

Count on t axis on graph

• Time taken=12s

#5

Just the initial velocity and height becomes zero

New function

• h(t)=-4.9t^2

#6

Sam e initial velocity but height becomes zero

New function

• h(t)=-4.9t^2+19.8t