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Kinematics - Projectile Motion

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A projectile is any object thrown into space upon which the only acting force is gravity. The path followed by a projectile is called its trajectory, which is a parabola.

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The motion of a projectile is a two-dimensional motion that can be resolved into two independent components: horizontal motion with constant velocity (since horizontal acceleration ax=0a_x = 0) and vertical motion with constant acceleration (acceleration due to gravity ay=−ga_y = -g).

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Assumptions made in projectile motion: There is no air resistance, the effect due to the rotation of Earth and its curvature is negligible, and the acceleration due to gravity gg is constant in magnitude and direction.

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Horizontal Projection: When a body is projected horizontally from a height hh with velocity uu, the time of flight is given by t=2hgt = \sqrt{\frac{2h}{g}} and the horizontal range is R=u2hgR = u \sqrt{\frac{2h}{g}}.

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Oblique Projection: When a body is projected with an initial velocity uu at an angle θ\theta with the horizontal.

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The horizontal range is the same for two angles of projection θ\theta and (90∘−θ)(90^\circ - \theta), provided the initial velocity uu remains the same.

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Velocity of the projectile at any time tt is given by the vector sum of its horizontal and vertical components: v=vx2+vy2v = \sqrt{v_x^2 + v_y^2} where vx=ucos⁡θv_x = u \cos \theta and vy=usin⁡θ−gtv_y = u \sin \theta - gt.

📐Formulae

ux=ucos⁡θ,uy=usin⁡θu_x = u \cos \theta, \quad u_y = u \sin \theta

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

T=2usin⁡θgT = \frac{2u \sin \theta}{g}

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

R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}

Rmax=u2gat θ=45∘R_{max} = \frac{u^2}{g} \quad \text{at } \theta = 45^\circ

v=(ucos⁡θ)2+(usin⁡θ−gt)2v = \sqrt{(u \cos \theta)^2 + (u \sin \theta - gt)^2}

💡Examples

Problem 1:

A cricketer can throw a ball to a maximum horizontal distance of 100 m100 \text{ m}. With the same effort, to what maximum vertical height can he throw the same ball? (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Given Rmax=100 mR_{max} = 100 \text{ m}. We know Rmax=u2gR_{max} = \frac{u^2}{g}. Therefore, u210=100  ⟹  u2=1000\frac{u^2}{10} = 100 \implies u^2 = 1000. For maximum vertical height, the ball must be thrown at θ=90∘\theta = 90^\circ. Using the formula H=u2sin⁡2θ2gH = \frac{u^2 \sin^2 \theta}{2g}, we get H=1000×(sin⁡90∘)22×10=1000×120=50 mH = \frac{1000 \times (\sin 90^\circ)^2}{2 \times 10} = \frac{1000 \times 1}{20} = 50 \text{ m}.

Explanation:

The maximum horizontal range is achieved at 45∘45^\circ, while the maximum vertical height is achieved when the projectile is thrown vertically upwards (90∘90^\circ). The maximum vertical height is exactly half of the maximum horizontal range for the same initial velocity.

Problem 2:

A body is projected with a velocity of 30 m/s30 \text{ m/s} at an angle of 30∘30^\circ with the horizontal. Find the time of flight and the horizontal range. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Initial velocity u=30 m/su = 30 \text{ m/s}, angle θ=30∘\theta = 30^\circ, g=10 m/s2g = 10 \text{ m/s}^2.

  1. Time of flight T=2usin⁡θg=2×30×sin⁡30∘10=60×0.510=3 sT = \frac{2u \sin \theta}{g} = \frac{2 \times 30 \times \sin 30^\circ}{10} = \frac{60 \times 0.5}{10} = 3 \text{ s}.
  2. Horizontal Range R=u2sin⁡2θg=302×sin⁡(2×30∘)10=900×sin⁡60∘10=90×32=453≈77.94 mR = \frac{u^2 \sin 2\theta}{g} = \frac{30^2 \times \sin(2 \times 30^\circ)}{10} = \frac{900 \times \sin 60^\circ}{10} = 90 \times \frac{\sqrt{3}}{2} = 45\sqrt{3} \approx 77.94 \text{ m}.

Explanation:

Calculated the total time the projectile remains in the air using the vertical component of motion and the total horizontal distance covered using the constant horizontal velocity component.

Projectile Motion - Revision Notes & Key Formulas | ICSE Class 11 Physics