Answer to Question #130991 in Physics for Kat

Question

1.      A cannonball is catapulated toward a castle. The cannon ball velocity when it leaves a catapult is 40 m/s at an angle of 37⁰ with respect to the horizontal and the cannonball is 7.0 m from the ground at this time. (a) What is the maximum height above the ground reach by the cannon ball? (b) What horizontal Distance from the release point will it land?

Answer 1

The maximum height is

$$H=h+\frac{(v\text{ sin}\alpha)^2}{2g}=7+\frac{(40\text{ sin}37°)^2}{2\cdot9.8}=37\text{ m}.$$

The horizontal distance is determined by the horizontal component of initial velocity and time of flight:

$$R=v\text{ cos}37°t.$$

The time of flight is made up of time t₁ required to go up from 7-m level, and then t₂ -from max. height to the ground:

$$t_1=\frac{v\text{ sin}\alpha}{g}=2.46\text{ s},\\\space\\ t_2=\sqrt{\frac{2H}{g}}=2.75\text{ s},\\\space\\ R=166\text{ m}.$$