Step 1: Identify the equation.
• When subbing in the appropriate values, the equation is h= -16t2+192t+32, with 192 being the initial velocity and 32 being the starting height.
Step 2: Identify your variables.
• A: -16
• B: 192
• C: 32
Step 3: Find the time (t)
• To find t you use the quadratic formula: -b±√(b2-4ac)/2a
• Sub in the values of a, b, and c as verified above.
• -(192)±√((1922)-(4*-16*32))/2(-16)
• This equation simplifies to -192±(√38912)/-32 = -.166 or 12.16
• Considering it is not possible to have a negative time, 12.16 is (t)≈12 seconds
Step 4: Find the height (h)
• To find the height, you must find first the x-coordinate of the vertex. This can be found in the equation x= -b/2a
• Sub in the above values to get -192/2(-16) which computes to 6.
• Next you have to simply sub in 6 for x (t) in the equation to get the y-coordinate (height)
h= -16t2+192t+32.
• h=-16(6)2+192(6)+32 which simplifies to h= -576+1152+32= 608 ft
Final Answer:
• Quadratic Model: h= -16t2+192t+32
• How high does the cannonball go? 608 ft
• How long is the cannonball in the air? 12 seconds
Your explanations of the steps were very excellent, elaborate, and concise.
ReplyDeletegood job Sammi :)
everything is explained really well and all makes sense, and your blog is pretty :)
ReplyDeleteGOOD JOB SAMMI! I like how your information all makes sense and is so straightforward! :D
ReplyDeleteOverall great job organizing your steps. It would've been better if you added the original formula to give a better reference. B)
ReplyDeleteI thimk that this is really good. I know I've already commented, but I think that it's really comming along and I LOVE how you described every single step in so much detail! Good Luck! ;D
ReplyDelete