For this part, we were to find the height of the cannon ball and how long it was in the air by using the formula: h = -16t2+v0t+h0. The H represents the final height, the -16 represents the pull of gravity, v0t represents the initial velocity, and h0 represents the initial height.
We were given a problem: “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.”
1.Finding the final height
1st. plug in our know units into the formula :
2nd. Find the value of t. t represents time, use the formula t=-b/(2a). plug in the units from the above equation.
3rd. t=(-192)/(2*-16) simply solve.
4th. t= 6 seconds
5th. then after you have the time, you simply insert it back into the original equation in step 1.
h=-16(6)2+192(6)+32 simply solve once more.
h=-576+1152+32r= range in legth
g= acceleration of gravity (meter/second^2)
however, this seems simpler: (speed (ft/sec))cos(launch angle)
I predict the trajectory will be smooth and the cannonball will shoot straight forward