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?

1 Answer
Sep 2, 2015

you just have to use equations of motion to solve this problem

Explanation:

enter image source here

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