A particle at a height h from the ground is projected with an angle 30 from the horizontal it strikes the ground making angle 45 with horizontal .It is again projected from the same pt at height h with the same speed but with an angle of 60 with horizontal.Find the angle it makes with the horizontal when it strikes the ground.
Asked by m.nilu
7th July 2018,
11:46 AM
Answered by Expert
Answer:
initial vertical distance h1 (refer figure )travelled by particle is obtained from " v2 = U2 - 2×g×h1 " ,
with final velocity v = 0 and initial velocity U = u×sin30 = u/2;
h1 = [ u2 × (1/4) ] /( 2 g ) = u2 / (8×g)
for reaching the ground, particle has to travel the vertical distance (h+h1) = h + [ u2 / (8×g) ]
vertical velocity component v1 of the particle while hitting the ground first time is obtained
from "v2 = 2×g×H ", because initial velocity is zero and H is the total distance.
v12 = 2×g×{ h + [u2 / (8×g) ] } = (u2/4 )+2×g×h .................(1)
it is given that particle hit the ground first time at an angle 45º.
Hence horizontal component of velocity and vertical component of velocity are same.
hence we have v2 = (u2/4 )+2×g×h = u2 cos230 = u2×(3/4) or u2 = 4×g×h .....................(2)
height h2 travelled when particle is projected second time = ( u2 × sin260 )/(2×g) = (3/2)×h ...............(3)
In eqn.(3), substitution is done for u2 from eqn.(2)
vertical distance travelled by projectile at second time, h+(3/2)h = (5/2)h
vertical veleocity component v2 of the particle, when it hit the ground second time, v22 = 2×g×(5/2)×h = 5×g×h
incident angle α for the particle when it is hitting the ground second time is given by, tan α = v2 / (u×cos60) = √5
hence angle α = tan-1 (√5) ≈ 66º
Answered by Expert
8th July 2018,
12:02 PM
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