You are shooting a ball out of a cannon into a bucket that is 3.25-m away. What angle should the cannon be pointed knowing that acceleration(due to gravity) is -9.8m/s^2, the cannon height is 1.8m, the bucket height is .26m and the flight time is .49s?

consider the above diagram i've drawn about the situation.
i have taken the angle of the canon as #theta#
since the initial velocity is not given, i will take it as #u#

the cannon ball is #1.8m# above the ground at the edge of the cannon as goes into a bucket which is #0.26m# high. which means the vertical displacement of the cannon ball is #1.8 - 0.26 = 1.54#

once you've figured this out, you just have to apply these data into the equations of motion.

considering the horizontal motion of the above scenario, i can write
#rarrs=ut#
#3.25 = ucos theta * 0.49#

#u= 3.25/(cos theta*0.49)#

for the vertical motion
#uarrs=ut+1/2at^2#

#-1.54=usintheta * 0.49 - 9.8/2 * (0.49)^2#

replace the #u# here by the expression we got from the previous equation

#-1.54=3.25/(cos theta*0.49)sintheta * 0.49 - 9.8/2 * (0.49)^2#

this is it. from here it's just the calculations you have to do..
solve the above expression for #theta# and that's it.

#-1.54=3.25 tan theta - 9.8/2 * (0.49)^2#

you will get an answer for #tan theta# from here. get the inverse value to get the magnitude of the angle #theta#