Question

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

Answer 1

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*1.7x10^3m/s`

V(h) = 975.08 m/s

Now, horizontal range is given as

`H = V(h) * t`

`H = 975.08 * t`

= 975.08t m .... (1)

b) Time of flight

Vertical component of velocity is given as

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

`V(v)=vsin55*1.7x10^3m/s`

V(v) = 1392.5585 m/s

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

`t=(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*284.19s\\H=277113.45m`