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Grade 11Mechanics

Q.Equation of trajectory of a particle in vertical plane is given as, y= -x^2+ 5x- 6.The particle is projected from point (2,0,0) and the XZ plane lies on the ground.Positive y-axis is along vertically upward direction.Then the range of particle will be
A)2 unit
B)3 unit
C)5 unit
D)1 unit

Profile image of jiten
8 Years agoGrade 11
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2 Answers

Profile image of sourav pal
8 Years ago
Ans, A) 2 unit
The range of the particle is along horizontal plane , means around x axis. As per question is mentioned projected from (2,0,0) so here defined x=2
Profile image of Venkat
8 Years ago
For the given projectile problem, y is zero at x = 2 and when x is maximum. Hence,
y=-x^{^{2}}+5x-6 \\@x=0; x^{^{2}}-5x+6=0 \\ (x-3)(x-2)=0
 
so y is zero at x = 2 or x =3. Since the projectile starts at 2, ignore the x = 2. The projectile ends at 3
Maximum range = Final point – starting point = 3-2 = 1 unit.
Hence, option (d) 1 unit