Answer:
a) H=277113.45m
b) t = 142.09 seconds
Solution
In this question we have given
initial speed, v= 1.7x10^3m/s
angle of projection with ground, α= 55 degree
a) Neglecting air resistance, horizontal component of velocity is given as
[tex]V(h)=vcos\alpha[/tex]............(1)
put value of v and α in equation(1)
[tex]V(h)=cos 55\times1.7x10^3m/s[/tex]
V(h) = 975.08 m/s
Now, horizontal range is given as
[tex]H = V(h) \times t[/tex]
[tex]H = 975.08 \times t[/tex]
= 975.08t m .... (1)
b) Time of flight
Vertical component of velocity is given as
[tex]V(v)=vsin\alpha[/tex]............(1)
[tex]V(v)=vsin55\times1.7x10^3m/s[/tex]
V(v) = 1392.5585 m/s
We know that time to reach at maximum height is given by formula
[tex]t=\frac{V(v)}{g}[/tex]
t= (1392.5585 m/s)/9.8m/s^2
time to reach at maximum height,t = 142.09seconds
Total time in air 2t = 284.19 seconds.
put value of t in equation (1)
Therefore horizontal range is given as
[tex]H=975.08m/s\times284.19s\\H=277113.45m[/tex]
