A particle is projected over a triangle from one extremity of its horizontal base. Grazing over the vertex, it falls on the other extremity of the base. If α and β be the base angles of the triangle and θ the angle of projection, then find the relation between the angles.
A particle is projected over a triangle from one extremity of its horizontal base. Grazing over the vertex, it falls on the other extremity of the base. If α and β be the base angles of the triangle and θ the angle of projection, then find the relation between the angles.
The correct option is D
Assume that the ball is thrown with a speed of u. Drop perpendicular AM to BC. Now by using trigonometry we can say
BM = htanα CM = htanβ
Now BM + CM = Range = 2u2sin θcos θg
⇒2u2 sinθ cosθg = htan α + htan β ------------(I)
Now let's look at the motion of the projectile from point B to A
Resolve the components and write equation of Motion
ux = u cosθ
ax = 0
Let's assume in time t the ball reaches the point A
⇒sx = BM = htanα ⇒ u cos θ = (htanα)1t --------------(II)
Similarly sy = uyt + 12at2
h = u sin θ t − 12 gt2 -------------(III)
Substituting the value of t from equation (II) in equation (III)
h = usinθ hu tanα cosθ − 12gh2u2‘tan2α cos2θ
⇒gh2u2tan2α cos2θ = tanθtanα − 1 = (tanθ − tanαtanα)
⇒ u2 = gh tanα2tan2α cos2θ(tanθ − tanα)
Substituting the above in equation (I)
2gh sinθ cosθ2 tan α cos2θ(tan θ − tan α)g = htanα + htanβ
tan ttan α(tan α(tanθ − tan α)) = tanβ + tanαtanα tanβ
1 − tan αtan θ = tan β(tan β + tan α)
tan β + tanα − tanβtan β + tan α = tan αtan θ
tan β + tan αtan α = tan βtan α
⇒ tan θ = tan α + tan β
Alternate method:
Let ABC be the triangle with base BC. Let h be the height of the vertex A, above BC. If AM be the perpendicular drawn on base BC from vertex A, then tanα = ha and tan β = hb.
Where BM = a and CM = b
Since A (a, h) lies on the trajectory of the projectile; y = x tan θ(1 − xR) -----------(i)
Therefore it should satisfy equation (i)
i.e., h = a tan θ(1 − aa+b) [∴Range R = a+b]ha = tan θ = h(a+bab)
⇒hb + ha = tan α + tan β
∴tan θ = tan α + tan β
Hence Proved.
Assume that the ball is thrown with a speed of u. Drop perpendicular AM to BC. Now by using trigonometry we can say
BM = htanα CM = htanβ
Now BM + CM = Range = 2u2sin θcos θg
⇒2u2 sinθ cosθg = htan α + htan β ------------(I)
Now let's look at the motion of the projectile from point B to A
Resolve the components and write equation of Motion
ux = u cosθ
ax = 0
Let's assume in time t the ball reaches the point A
⇒sx = BM = htanα ⇒ u cos θ = (htanα)1t --------------(II)
Similarly sy = uyt + 12at2
h = u sin θ t − 12 gt2 -------------(III)
Substituting the value of t from equation (II) in equation (III)
h = usinθ hu tanα cosθ − 12gh2u2‘tan2α cos2θ
⇒gh2u2tan2α cos2θ = tanθtanα − 1 = (tanθ − tanαtanα)
⇒ u2 = gh tanα2tan2α cos2θ(tanθ − tanα)
Substituting the above in equation (I)
2gh sinθ cosθ2 tan α cos2θ(tan θ − tan α)g = htanα + htanβ
tan ttan α(tan α(tanθ − tan α)) = tanβ + tanαtanα tanβ
1 − tan αtan θ = tan β(tan β + tan α)
tan β + tanα − tanβtan β + tan α = tan αtan θ
tan β + tan αtan α = tan βtan α
⇒ tan θ = tan α + tan β
Alternate method:
Let ABC be the triangle with base BC. Let h be the height of the vertex A, above BC. If AM be the perpendicular drawn on base BC from vertex A, then tanα = ha and tan β = hb.
Where BM = a and CM = b
Since A (a, h) lies on the trajectory of the projectile; y = x tan θ(1 − xR) -----------(i)
Therefore it should satisfy equation (i)
i.e., h = a tan θ(1 − aa+b) [∴Range R = a+b]ha = tan θ = h(a+bab)
⇒hb + ha = tan α + tan β
∴tan θ = tan α + tan β
Hence Proved.
