A cannonball is fired in such a way that its horizontal range is equal to 3
times its maximum height. Find the
launch angle of this cannonball.

θ = 53.13° above horizontal

Explanation:

Ignore air resistance.

The time the ball takes to reach its max height will equal the time to travel half the max range R.

√(2h/g) = (R/2)/vcosθ

vsinθ/g = R/2vcosθ

2v²sinθcosθ = Rg

2v²sinθcosθ = 3hg

v² = u² + 2gh

0² = v²sin²θ + 2gh

v²sin²θ = 2gh

v² = 2gh/sin²θ

               2v²sinθcosθ = 3hg

(2gh/sin²θ)2sinθcosθ = 3hg

            (2/sinθ)2cosθ = 3

                            cotθ = 3/4

                                 θ = 53.130

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.