Question

Please answer and explain the equations of how you got the answer

Answer 1

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

$V(h)=vcos\alpha$............(1)

put value of v and α in equation(1)

$V(h)=cos 55\times1.7x10^3m/s$

V(h) = 975.08 m/s  

Now, horizontal range is given as

$H = V(h) \times t$

$H = 975.08 \times t$

= 975.08t m .... (1)  

b) Time of flight

Vertical component of velocity is given as

$V(v)=vsin\alpha$............(1)

$V(v)=vsin55\times1.7x10^3m/s$

V(v) = 1392.5585 m/s  

We know that time to reach at maximum height is given by formula

$t=\frac{V(v)}{g}$

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

$H=975.08m/s\times284.19s\\H=277113.45m$