A cannon with its open end at an initial height of 1 meter above the ground fire

Question

A cannon with its open end at an initial height of 1 meter above the ground fires a cannonball at an angle of 40° above the horizontal. It travels a horizontal distance of 40.0 meters to a 4 meter high wall, and passes over 3 meters above the wall. (Ignore air resistance)

a. Determine how much time it takes for the ball to reach the wall.
b. Determine the initial speed of the cannonball as it is fired.
c. Determine the velocity (magnitude and direction) of the cannonball as it passes over the castle wall.
d. If the same cannonball is fired with the same initial speed at an angle of 55.5° above the horizontal, determine whether (and by how much) the ball will still clear the castle wall.

Explanation / Answer

(b)initial height=1 m

final height=4+3=7 m

horizontal distance travelled=40 m

let the velocity be v

initial horizontal velocity=vcos40

initial vertical velocity=vsin40

using Newton equations

vcos(40)*t=40

6=vsin(40)*t-0.5*9.8*t^2

putting t=40/vcos(40)

we got v=22.01 m/s

(a)t=40/vcos40

we got t=2.37 sec

(c)after t=2.37 sec

vertical velocity=22.01sin 40 -9.8*2.37=-9.07 m/s

horizontal velocity=vcos(40)=16.86 m/s

final velocity at wall=sqrt(9.07^2+16.86^2)=18.34 m/s

direction=28.87 degree downwards

(d)if theta=55.5 degree

t=time taken to cover 40 m =40/vcos(55.5)=3.2 sec

distance travelled vertically=vsin(55.5)*t-4.9*t^2=7.86 m

so it will cross by 3.86 m

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