Question

A cannon, located 44.2m from the base of a vertical 29.0m tall cliff, shoots a 15.0kg shell at 40.8° above the horizontal toward the cliff.

What must the minimum muzzle velocity be for the shell to clear the cliff edge?

The ground at the top of the cliff is level, with a constant elevation of 29.0m above the cannon. How far past the cliff edge does the shell land when the shell is fired at that minimum speed?

What is the magnitude of the acceleration of the shell just before it hits the ground?

For this launch angle, what is the furthest distance to the cliff\'s base that the cannon can be moved to and still reach the level ground at the cliff\'s top given this launch angle and velocity?

Answer 1

given are

s1=44.2m

s2=29m

m=15kg

theta=40.8 degrees

x and y component of velocity is

vox=vcos(40.8)

Voy=vsin(40.8)

S1=voxt

44.2=vcos(40.8)t

T=44.2/vcos(40.3)

Y=voyt+1/2at^2

29=vsin(40.3)t+1/2(-9.8)t^2

29=vsin(40.3)t-4.9t^2

29=vsin(40.3)(44.2/vcos40.3)-4.9(44.2/vcos(40.3)^2

4.9*1953.6/v^2cos(40.3)=44.2tan(40.3)-29

   12551.5/v^2=8.48

12551.5/8.48=v^2

V=38.4m/sec—answer

T=44.2/vcos(40.3)

=44.2/38.4cos(40.3)

T=1.5 sec

Using above equation

29=vsin(40.3)t-4.9t^2

29=38.4sin(40.3)t-4.9t^2

4.9t^2-38.4 sin(40.3)t+29=0

4.9t^2-24.8t+29=0

T1=3.22sec

T2=1.8sec

X=voycos(40.3)t

=38.4 cos (40.3)*1.8

x=52.7 m