Question : At what angle should a body be projected with a velocity of 24 m/s just to pass over the obstacle 16 m high at a horizontal distance of 32 m. [Take g=10 m/s²]
Doubt by Jaskirat
Solution :
u = 24 m/s
x = 32 m
y = 16 m
Consider the horizontal distance covered by the body
s=ut+½at2 x=uxt+½axt2 x=ucosθ×t+½(0)t2 x=ucosθ×t+0 x=ucosθ×t x/ucosθ = t t=x/ucosθ t=32/24cosθ t=4/3cosθ — (1)
Now consider the vertical motion of the projectile
s=ut+½at2
y=uyt+½ayt2
y=usinθt+½(-g)t2
16=24sinθt+½(-10)t2
16=24sinθt-5t2
16=24sinθ(4/3cosθ)-5(4/3cosθ)2
[From equation (1)]
16=[(24×4)/3][sinθ/cosθ]-[(5×16)/9][1/cos2θ]
16=32tanθ-(80/9)sec2θ [∵1/cosθ = secθ]
Multiply both sides by 9
144=288tanθ-80sec2θ
144=288tanθ-80(1+tan2θ)
144=288tanθ-80-80tan2θ
144+80=288tanθ-80tan2θ
224=288tanθ-80tan2θ
80tan2θ-288tanθ+224=0
16(5tan2θ-18tanθ+14) = 0
5tan2θ-18tanθ+14 = 0
Let tanθ = z 5z2-18z+14 =0 Solving the quadratic equation by using the Quadratic Formula.
a=5 b=-18 c=14
D=b2-4ac D=(-18)2-4(5)(14) D= 324-280 D=44
z=[-b±√D]/2a z=[-(-18)±√44]/2(5) z=[18±6.63]/10 z=18/10±6.63/10 z=1.8±0.663 z=1.8+0.663 OR z=1.8-0.663 z=2.463 OR z=1.137
But z=tanθ
tanθ=2.463 OR tanθ=1.137 θ=tan-1(2.463) OR θ=tan-1(1.137) θ=67.90 OR θ=48.66
Similar Question : At what angle should a body be projected with a velocity 24 m/s just to pass over the obstacle 14 m high at a horizontal distance of 24 m. [Take g=10 m/s2] a) tanθ=3.8 b) tanθ=1 c) tanθ=3.2 d) tanθ=2