Question : A body is projected with a velocity of 40 m/s. After 2 sec it crosses a vertical pole of height 20.4 m. Calculate the angle of projection and horizontal range.
Doubt by Aakarsh
Solution :
Initial velocity (u) = 40 m/s
At t = 2sec
Vertical Distance Covered (y) = 20.4 m
We know,
At t = 2sec
Vertical Distance Covered (y) = 20.4 m
We know,
uy=usinθ
Using Equation
S=ut+½at²
y=uy(t)+½(-g)t²
20.4=usinθ(2)-½(9.8)(2)² [∵uy=ucosθ]
20.4=40sinθ(2)-(9.8)(2)
20.4=80sinθ-19.6
20.4+19.6=80sinθ
40=80sinθ
40/80=sinθ
S=ut+½at²
y=uy(t)+½(-g)t²
20.4=usinθ(2)-½(9.8)(2)² [∵uy=ucosθ]
20.4=40sinθ(2)-(9.8)(2)
20.4=80sinθ-19.6
20.4+19.6=80sinθ
40=80sinθ
40/80=sinθ
1/2=sinθ
sin30°=cosθ
sin30°=cosθ
θ=30°
Horizontal Range
=u²sin2(θ)/g
=(40)²sin2(30°)/9.8
=1600sin60°/9.8
=1600(√3/2) / 9.8
= 800√3/9.8
= 141.39 m
=u²sin2(θ)/g
=(40)²sin2(30°)/9.8
=1600sin60°/9.8
=1600(√3/2) / 9.8
= 800√3/9.8
= 141.39 m
Hence, the angle of projection is 30° and Horizontal Range is 141.39 m.