Asked by john
                a ball that was projected up at an angle of θ degrees with the horizontal has a flight time of 5.0s and covers a range of 49.0 m. find (a) the time it took to reach maximum height, (b) the horizontal and vertical components of its initial velocity, (c) its initial velocity,(d) the angle θ at which it was projected, and (e) the maximum height it reached.
            
            
        Answers
                    Answered by
            Damon
            
    vertical problem:
2.5 seconds up and 2.5 seconds down
v = Vi - 9.81 t
at top v = 0
so
Vi = 9.81*2.5 = 24.5 m/s up
h = Vi t - 4.9 t^2
= 24.5(2.5) -4.9(6.25)
= 61.3 - 30.6
= 30.6 meters high at top
horizontal problem
49 = u (5)
so u = 9.8 m/s
tan theta = Vi/u = 24.5/9.8
theta = 68.2 degrees
    
2.5 seconds up and 2.5 seconds down
v = Vi - 9.81 t
at top v = 0
so
Vi = 9.81*2.5 = 24.5 m/s up
h = Vi t - 4.9 t^2
= 24.5(2.5) -4.9(6.25)
= 61.3 - 30.6
= 30.6 meters high at top
horizontal problem
49 = u (5)
so u = 9.8 m/s
tan theta = Vi/u = 24.5/9.8
theta = 68.2 degrees
                    Answered by
            Anonymous
            
    the horizontal and vertical component of its initial velocity of 5.0s and 49.0m
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.