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Question 28 of 28

Q.a) A particle is projected with an initial velocity u, making an angle θ with the horizontal. Find the equation of the trajectory of the particle at any instant after the projection. Find expressions for the maximum height gained by the particle and its horizontal range. b) What are the quantities that remain constant during the motion of the particle? OR

a) What is meant by relative velocity? A particle A is moving with a velocity u. Another particle B is moving with a velocity v in a direction making an angle θ with the direction of motion of the first particle. Find an expression of the velocity of the second particle relative to the velocity of the first particle. b) Draw the velocity-time graph of a particle moving with non-uniform acceleration. From the graph find the instantaneous and average acceleration of the particle.
West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 5mImportance★★★★★
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Eliminate time between the horizontal and vertical equations of motion to get the parabolic trajectory, then use it (or standard results) for maximum height and range.

a) Trajectory equation: A particle projected with speed uu at angle θ\theta has:

x=(ucos⁡θ)t  ⟹  t=xucos⁡θx=(u\cos\theta)t \implies t=\frac{x}{u\cos\theta}

y=(usin⁡θ)t−12gt2y=(u\sin\theta)t-\tfrac12 gt^2

Substituting tt:

y=xtan⁡θ−gx22u2cos⁡2θy=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}

This is the equation of a parabola (the trajectory of projectile motion).

Maximum height: at the highest point, vertical velocity component =0=0:

H=u2sin⁡2θ2gH=\frac{u^2\sin^2\theta}{2g}

Horizontal range: the horizontal distance covered when the particle returns to the same height (time of flight T=2usin⁡θgT=\dfrac{2u\sin\theta}{g}):

R=ucos⁡θ⋅T=u2sin⁡2θgR=u\cos\theta\cdot T=\frac{u^2\sin2\theta}{g}

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