A cannonball is shot upward from the upper deck of a fort with an initial velocity of 192 feet per second. The deck is 32 feet above the ground.
Use this formula to solve the problem: h = -16t2+v0t+h0
Use this formula to solve the problem: h = -16t2+v0t+h0
Quadratic Model: -16t2+192t+32
1. How high does the cannonball go? 608 feet (Remember you are looking for a specific part of the vertex.)
2. How long is the cannonball in the air? 12.17 seconds (Remember you can use the quadratic formula.)
Steps for finding th height:
· Plug in our know units into the formula :
h=-16t2+192t+32
· Find the value of t. t represents time, use the formula t= -b/(2a).
· Now plug in the units from the equation above.
t=(-192)/(2*-16) Solve.
t= 6 seconds
· Now that you have the time, you can simply substitute it back into the original equation in the first step.
h=-16(6)2+192(6)+32
h=-576+1152+32
h=608 feet.
Steps for Finding the time in the air:
· Plug in the known units into the formula:
h=-16t2+192t+32
· Plug your units into the quadratic formula which is:
-b ±√b2-4ac
2a
· Plug in our know units: -192±√(192)2-4(-16)(32)
2 (-16)
· When you solve what is inside the square root symbol and you should end up with this:
-192±√38912
-32
· Find the square root of what is in the inside the square root:
-192 ± 197.3
-32
· Now divide -32 into the top numbers and you recieve:
6 ± -6.17
· Since we are solving for time; you obviously can’t have a negative time, so you must solve for the answer which is positive. And that answer should be:
t=12.17 seconds
Great job, very easy to follow!
ReplyDeleteGreat Math!
ReplyDeleteGreat job explaining the process you made it very easy to follow
ReplyDelete