# 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?

Sep 2, 2015

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

#### Explanation:

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 $1.8 m$ above the ground at the edge of the cannon as goes into a bucket which is $0.26 m$ 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
$\rightarrow s = u t$
$3.25 = u \cos \theta \cdot 0.49$

$u = \frac{3.25}{\cos \theta \cdot 0.49}$

for the vertical motion
$\uparrow s = u t + \frac{1}{2} a {t}^{2}$

$- 1.54 = u \sin \theta \cdot 0.49 - \frac{9.8}{2} \cdot {\left(0.49\right)}^{2}$

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

$- 1.54 = \frac{3.25}{\cos \theta \cdot 0.49} \sin \theta \cdot 0.49 - \frac{9.8}{2} \cdot {\left(0.49\right)}^{2}$

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

$- 1.54 = 3.25 \tan \theta - \frac{9.8}{2} \cdot {\left(0.49\right)}^{2}$

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