Answer to Question #189714 in Mechanics | Relativity for Fhulu

Question

A cannon ball is catapulted towards the castle. The cannonball's velocity when it leaves the catapult is 40 m/s at an angle of 37° with respect to the horizontal, and the cannonball is 7 m above the ground at this time. a) Calculate the maximum height above the ground reached by the cannonball. b) Assuming the cannonball makes it over the castle walls and lands down on the ground, calculate at what horizontal distance from its release point will it land. c) Calculate the x- and y- components of the cannonball's velocity just before it hits the ground. [Hint : y-axis points up]

Answer 1

Solution :-

U = 40 m/s

$\theta = 37^o$

$V_x = Ucos\theta =32m/s \\ V_y=Usin\theta = 24 m/s$

g = 9.8m/s²

(1) The maximum height calcualated by

$$H=h+\frac{(U\text{ sin}\theta)^2}{2g}=7+\frac{(40\text{ sin}37°)^2}{2\cdot9.8}=36.3\text{ m}.$$

(2)

The horizontal distance is

we can calculate this using horizontal component of velocity and time of flight

$$R=U\text{ cos}37°t.$$

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:

$t_1=\frac{U\text{ sin}\theta}{g}=2.45\text{ s},\\\space\\ t_2=\sqrt{\frac{2H}{g}}=2.72\text{ s},\\\space\\ R=165.5\text{ m}.$

(3)

x component of U will remain same but y component will change due to acceleration due to gravity

so final velocity in y direction just hitting ground is

$= 24 - 9.8\times5.17 \\ =-26.6 m/s$ ( using V = u +at formula)

velocity in x direction before hitting ground = 32 m/s

so finally the function

$V_f=32 x- 26.6y$