A cannon ball is catapulted toward a castle. The cannon's ball velocity when it leaves the catapult is 40m/s

Question

at an angle of 37 degrees w/ respect to the horizontal and the cannonball is 7m above the ground at this time.
a. What is the maximum height above the ground reached by the cannonball?
b. Assuming the castle walls and lands backdown on the ground, at what horizontal distance from its release point will it lands?The y-axis points up.

Answer 1

First, construct some equations of motion for the cannon ball:
In the vertical
vy(t)=40*sin(37)-9.8*t
y(t)=7+40*sin(37)*t-.5*9.8*t^2

In the horizontal
vx(t)=40*cos(37)
x(t)=40*cos(37)*t

a) Max height, or apogee, occurs when vy(t)=0
0=40*sin(37)-9.8*t
solve for t
t=2.4564
y(2.4564)=36.54 m

b) when y(t)=0, for t>0, the cannon ball strikes the ground
0=7+40*sin(37)*t-.5*9.8*t^2
solve for t
t=5.188
x(5.188)=165.74

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Answer 2

on your individual cosmic gag reel, a catapult. A cannon is going boooom. A catapult you spot the *ingredient* flinging flow.... woooosh Now that sounds in simple terms extra powerful, and then your hear the fluuuuubbbshh .....while she hits the floor.