Question

A particle is projected so as to graze tops of 2 samll towers each of height 10 m at distance of 100m and 20 m from point of projection. What will be horizontal range, max. Height , time of flight, angle at which it is launched

Answer 1

Answer:

solved

Explanation:

angle of projection = ∝ ,  u = initial speed of projection , g = 10m/s²

x = utcos∝ , y = utsi∝ - 1/2gt² , t = xsec∝/u

y = xtan∝ - 5x²sec²∝/u²

assuming the coordinate of the point of projection to be (0,0) , the trajectory passes through (20,10)  ,  (120,10)

10 =  120 tan∝ - 5(120)²sec²∝/u² = 20tan∝ - 5(20)²sec²∝/u²

100 tan∝ =  70000sec²∝/u² , tan∝/700 = sec²∝/u²

20tan∝ - 20/7 tan∝ = 10 , tan∝ = 7/12 , cos ∝ = 12/√189 , sin ∝ = 7/√189

 u² /sec²∝ = 700/tan∝ , u² = 700sec²∝/tan∝ = 700= 700/sin∝cos∝

u² = 700 ÷ 84/189 = 225x7 , u = 15√7 m/sec

Now

horizontal range   = u²sin2∝/g =   225x7 x 2x84/1890 = 140m.

max. Height  = u²sin²∝/2g = 225x7x49/(1890x2) = 20.417 m

time of flight  =2u sin∝/g = 3√7 x7/3√21 = 7√3/3 sec.

angle at which it is launched = ∝ = tan∧-1 (7/12  ) = 30.26°