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Various Parameters Of Projectile

Practice to find maximum height of a projectile, vertical displacement of a projectile & velocity of a particle at a given instant. Learn formulas how to calculate of projectile motion equations.

Position of a Particle at Given Instant

  • A particle is projected with an initial velocity u, making an angle \(\theta\) with the horizontal, as shown in figure.
  • Point of projection of particle is considered as origin.

        Time of Flight 

         Time of flight of a particle projected along a horizontal plane

          \(T=\dfrac {2\,u\,sin\theta}{g}\)

       Displacement along \(x\;and\ y\) axis at time \((t<T)\)

Here, t < T (Time of flight)

Parameters along x-axis

  • Initial velocity, \(u_x=u\,cos\,\theta\)
  • Acceleration, \(a_x=0\)
  • Displacement along x-axis at time t, x(t)

using  \(s=ut+\dfrac {1}{2}at^2\)

        \(x(t)=u_xt+\dfrac {1}{2}a_xt^2\)

        \(x(t)=u_xt\)

         \(x(t)=(u\,cos\,\theta)\,t\)

Parameters along y-axis

  • Initial velocity, \(u_y=u\,sin\,\theta\)
  • Acceleration, \(a_y=-g\)
  • Displacement along y-axis at time 't', y(t) using

          \(s=ut+\dfrac {1}{2}at^2\)

           \(y(t)=u_yt+\dfrac {1}{2}a_yt^2\)

          \(y(t)=(u\,sin\,\theta)\,t-\dfrac {1}{2}gt^2\)    

The acceleration along \(x\;and \;y\) axis are shown in figure.

Illustration Questions

A particle is projected with velocity \(u=20\,m/s\) , making an angle \(\theta=30°\) with the horizontal. Determine position \(x\) and \(y\) at time \(t=1\,sec\).  Assuming point of projection as origin. [ Take \(g=10\,m/s^2\) ]

A \(x=10\sqrt 3\,m,\,\;y=5\,m\)

B \(x=10\,m,\,\;y=5\,\sqrt 3\,m\)

C \(x=5\,m,\,\;y=5\,m\)

D \(x=4\,m,\,\;y=3\,m\)

×

Given : Initial velocity, \(u=20\,m/s\) , \(\theta=30°\) 

image

Parameters along x-axis

\(u_x=u\,cos\,30°\)

     \(=20.\dfrac {\sqrt 3} {2}\)

\(u_x=10\sqrt 3\;m/s\)

\(a_x=0\)

Parameters along y-axis

\(u_y=u\,sin\,30°\)

\(u_y\) \(=20×\dfrac {1} {2}\)

\(u_y=10\;m/s\)

\(a_y=-g\) 

image image

Position of a particle at time t,

using  \(s(t)=ut+\dfrac {1}{2}at^2\)

\(x(t)=u_xt+\dfrac {1}{2}a_xt^2\)           (at  \(t=1\,sec\) )

\(x(1)=10\sqrt 3×1+\dfrac {1}{2}×0×(1)^2\)

\(\Rightarrow x(1)=10\sqrt 3\;m\)

 

\(y(t)=u_yt+\dfrac {1}{2}a_yt^2\)

\(y(1)=10×1+\dfrac {1}{2}×(-g)×(1)^2\)

\(\Rightarrow y(1)=10-5\;\)

\(\Rightarrow y(1)=5\;m\)     

image image

A particle is projected with velocity \(u=20\,m/s\) , making an angle \(\theta=30°\) with the horizontal. Determine position \(x\) and \(y\) at time \(t=1\,sec\).  Assuming point of projection as origin. [ Take \(g=10\,m/s^2\) ]

image
A

\(x=10\sqrt 3\,m,\,\;y=5\,m\)

.

B

\(x=10\,m,\,\;y=5\,\sqrt 3\,m\)

C

\(x=5\,m,\,\;y=5\,m\)

D

\(x=4\,m,\,\;y=3\,m\)

Option A is Correct

Velocity of a Particle at Given Instant

  • A particle is projected with an initial velocity u, making an angle \(\theta\) with the horizontal, as shown in figure.
  • Velocity of a particle at any time is v \((t<T)\), where T = time of flight.

 

Parameters along x-axis

  • Initial velocity, \(u_x=u\,\cos\theta\)
  • Acceleration \(a_x=0\)
  • Velocity along x-axis at time t,

           using \(v=u+at\)

          \(v_x=u_x+a_xt\)

           \(v_x\)  \(=u_x+0.t\)

           \(v_x=u_x\)

           \(v_x=u\,cos\,\theta\)

Parameters along y-axis

  • Initial velocity, \(u_y=u\,\sin\theta\)
  • Acceleration \(a_y=-g\)
  • Velocity along y-axis at time t,

           using \(v=u+at\)

          \(v_y=u_y+a_yt\)

         \(v_y=u_y-g\,t\)

         \(v_y=u\,sin\,\theta-g\,t\)     

Velocity at time t

\(\vec v=v_x\hat i+v_y\hat j\)

\(\vec v=u_x\hat i+(u_y-g\,t)\,\hat j\)

\(|\vec v|=\sqrt {v_x^2+v_y^2}\)    

Illustration Questions

A particle is projected with velocity \(u=20\,m/sec\)  making an angle \(\theta=30°\) with the horizontal. Determine its velocity at time \(t=1\,sec\). [ Take \(g=10\,m/s^2\)]

A \(5\, m/s\)

B \(4\, m/s\)

C \(10\sqrt 3\, m/s\)

D \(3\, m/s\)

×

Given:

Initial velocity \(u=20\, m/s\)\(\theta=30°\)

image

Parameters along x-axis

\(u_x=u\,cos\,30°\)

\(u_x\) \(=20×\dfrac {\sqrt 3}{2}\)

\(u_x=10\sqrt 3\,m/s\)

\(a_x=0\)

Parameters along y-axis

\(u_y=u\,sin\,30°\)

      \(=20×\dfrac {1}{2}\)

\(u_y=10\,m/s\)

\(a_y=-g\)   

image image

Velocity of particle at time t = 1 sec

using

\(v=u+at\)

\(v_x(t)=u_x+a_xt\)

\(v_x(1)=10\,\sqrt 3\,m/sec\)

 

 

\(v_y(t)=u_y+a_y.t\)

\(v_y(1)=10-10 \)

\(v_y(1)=0\,m/sec\)

 

Velocity

\(\vec v=v_x\hat i+v_y\hat j \)

\(\vec v=10\sqrt 3\,\hat i\)

\(|\vec v|=10\sqrt 3\,m/sec\)

 

 

image

A particle is projected with velocity \(u=20\,m/sec\)  making an angle \(\theta=30°\) with the horizontal. Determine its velocity at time \(t=1\,sec\). [ Take \(g=10\,m/s^2\)]

image
A

\(5\, m/s\)

.

B

\(4\, m/s\)

C

\(10\sqrt 3\, m/s\)

D

\(3\, m/s\)

Option C is Correct

Kinetic Energy of a Particle at Highest Point

  • A particle is projected with an initial velocity u, making an angle \(\theta\) with the horizontal, as shown in figure.

Parameters along x-axis

  • Initial velocity  \(u_x=u\,cos\,\theta\)
  • Acceleration  \(a_x=0\)

Parameters along y-axis

  • Initial velocity \(u_y=u\,sin\,\theta\)
  • Acceleration \(a_y=-g\)   

At Highest point

  • Velocity of a particle in y-direction \(v_y=0\)
  • Velocity of a particle in x-direction \(v_x=u_x=u\,cos\,\theta\)
  • Speed of particle at highest point \(=u\,cos\,\theta\)

As we know, 

Kinetic Energy = \(\dfrac {1}{2}\,m\,v^2\)

Kinetic Energy(At highest point) = \(\dfrac {1}{2}\,m\;(u\,cos\,\theta)^2\)    

Illustration Questions

A particle is projected with velocity u = 20 m/s, making an angle \(\theta\) = 37° with the horizontal. If mass of the particle is 2 kg then determine its momentum at the highest point. [Take \(sin\,37°=\dfrac {3}{5}\,,\,cos\,37°=\dfrac {4}{5}\)]

A 16 N sec

B 20 N sec

C 15 N sec

D 32 N sec

×

Given:

Initial velocity, u = 20 m/s, \(\theta\)= 37°

image

Parameters along x-axis

\(u_x=u\,cos\,37°\)

\(=20×\dfrac {4}{5}\)

\(=16\, m/s\)

\(a_x=0\)

Parameters along y-axis

\(u_y=u\,sin\,37°\)

\(=20×\dfrac {3}{5}\)

\(=12\, m/s\)

\(a_y=-g\)   

image image

Momentum at highest point

\(|\vec P|=m\cdot v_x\) (Because at highest point \(\vec {v_y}=0\) and \(v_x=u_x\))

       \(=2×16\)

\(|\vec P|= 32\;N\,sec\)   

image

A particle is projected with velocity u = 20 m/s, making an angle \(\theta\) = 37° with the horizontal. If mass of the particle is 2 kg then determine its momentum at the highest point. [Take \(sin\,37°=\dfrac {3}{5}\,,\,cos\,37°=\dfrac {4}{5}\)]

image
A

16 N sec

.

B

20 N sec

C

15 N sec

D

32 N sec

Option D is Correct

Calculation of Vertical Displacement of a Projectile at a Given Instant

  • A particle is projected with an initial velocity u, making an angle \(\theta\) with the horizontal, as shown in figure.
  • Time of Flight

    Time of Flight  \(T=\dfrac {2\,u\,sin\,\theta}{g}\)

  • To calculate displacement along y-axis or vertical displacement at  \(t<T\).
  • Initial velocity along y-direction \(v_y=u\,sin\;\theta\)
  • Acceleration \(a_y=-g\)
  • Displacement using \(s(t)=ut+\dfrac {1}{2}at^2\)
  • \(y(t)=u\,sin\,\theta\,.t-\dfrac {1}{2}\,g\,t^2\)    

Illustration Questions

A particle is projected with a velocity \(u = 50\; m/sec\)  making an angle \(\theta = 30°\) with the horizontal. Calculate its vertical displacement at time \(t = 3\; sec\).

A 10 m

B 20 m

C 15 m

D 30 m

×

Given:

Initial velocity \(u = 50\; m/sec\) , \(\theta = 30°\) 

image

Initial velocity along y-direction

\(v_y=u\;sin\;\theta\)

    \(=50\;sin\;30°\)

\(=50×\dfrac{1}{2}\)

\(u_y=25\;m/sec\)

Acceleration, \(a_y\) = – g

image image

Vertical displacement at t = 3 sec

\(y(t)=u_yt-\dfrac {1}{2}\;gt^2\)

\(=25×3-\dfrac {1}{2}\;×10×(3)^2\)

\(=75-45\)

\(=30\;m\)

image image

A particle is projected with a velocity \(u = 50\; m/sec\)  making an angle \(\theta = 30°\) with the horizontal. Calculate its vertical displacement at time \(t = 3\; sec\).

A

10 m

.

B

20 m

C

15 m

D

30 m

Option D is Correct

Calculation of Time at which the Projectile is at Height less than Maximum Height

Maximum Height

  • A particle is projected with initial velocity \(u\), making an angle \(\theta\) with the horizontal.

          \(H_{max}=\dfrac {u^2\;sin^2\;\theta}{2g}\)

Motion in y-direction (Vertical Displacement)

  • Initial velocity in y-direction 

           \(u_y=u\;sin\;\theta\)

          \(a_y=-g\)

 

Vertical Displacement, at time t 

 using \(s=ut+\dfrac {1}{2}\,at^2\)       \((\because\,\, h=s)\)

\(h=u\;sin\;\theta \cdot t-\dfrac {1}{2}\,gt^2\)

\(gt^2-2\;u\;sin\;\theta\cdot t+h=0\)

\(t=\dfrac {2\;u\;sin\;\theta\pm \sqrt {4\;u^2\;sin^2\;\theta-4gh}}{2g}\)

 

Here, two values of time indicating projectile is achieving same value of height h, for both the instants.

Illustration Questions

A particle is projected with a velocity \(u = 78\; m/s\), making an angle \(\theta = 23°\)with the horizontal. Calculate the time when the particle is at height \(h = 40\; m\).\(\Big [Take \,\,sin\;23°=\dfrac {5}{13}\Big]\)

A \(t=4\;sec,\;2\;sec\)

B \(t=5\;sec,\;8\;sec\)

C \(t=3\;sec,\;8\;sec\)

D \(t=8\;sec,\;4\;sec\)

×

Given:

Initial velocity \(u = 78\; m/s\), \(\theta = 23°\) and  \(h = 40\; m\)

image

\(u_y=u\sin\;\theta\)

     \(=75×sin\,23°\)

\(=78 ×\dfrac{5}{13}\)

    \(=30\;m/sec\)

\(a_y=-g\)

\(a_y=-10\;m/s^2\)

image image

Use

\(s=ut+\dfrac {1}{2}\,at^2\)

\(\Rightarrow h=u_yt-\dfrac {1}{2}\;gt^2\)

\(\Rightarrow 40=30t-5t^2\)

\(\Rightarrow5t^2-30t+40=0\)

\(t=\dfrac {30\pm\sqrt {900-4×40×5}}{2×5}\)

\(t=\dfrac {30\pm10}{10}\)

\(t=\dfrac {40}{10},\;\dfrac {20}{10}\)

\(t=4\;sec,\;2\;sec\)

image

A particle is projected with a velocity \(u = 78\; m/s\), making an angle \(\theta = 23°\)with the horizontal. Calculate the time when the particle is at height \(h = 40\; m\).\(\Big [Take \,\,sin\;23°=\dfrac {5}{13}\Big]\)

A

\(t=4\;sec,\;2\;sec\)

.

B

\(t=5\;sec,\;8\;sec\)

C

\(t=3\;sec,\;8\;sec\)

D

\(t=8\;sec,\;4\;sec\)

Option A is Correct

Illustration Questions

A particle is projected with velocity \(u = 20\; m/sec\) making an angle of \(\theta = 37°\) from a height of \(h = 20\; m\), as shown in figure. Calculate the time when final velocity is perpendicular to the initial velocity.

A \(t=\dfrac {10}{3}\;sec\)

B \(t=\dfrac {20}{3}\;sec\)

C \(t=4\;sec\)

D \(t=2\;sec\)

×

Given,

Initial velocity \(u = 20\; m/sec\)\(\theta = 37°\)\(h = 20\; m\)

image

Initial velocity,

\(\vec u=u_x\hat i+u_y\hat j\)

\(=(u\;cos\;\theta)\hat i+(u\;sin\;\theta)\hat j\)

\(=(20\;cos\;37°)\hat i+(20\;sin\;37°)\hat j\)

\(=\left (20×\dfrac {4}{5}\right)\hat i+ \left (20×\dfrac {3}{5}\right)\hat j\)

\(\vec u=16\;\hat i+12\;\hat j\)

image

Final velocity at time t,

Use \(v=u+at\)

\(\vec v=\vec u+\vec {a_y}\;t\)

\(\vec v=(u_x\hat i+u_y\hat j)-g\;t\;\hat j\)

\(\vec v=16\;\hat i+(12-10t)\;\hat j\)

image

For \(\vec v \perp\vec u\)

    \(\\\;\vec u.\vec v=0\)

image

\((16\;\hat i+12\;\hat j)\cdot [16\;\hat i+(12-10t)\;\hat j]=0\)

\((16)^2+(12)^2-120\;t=0\)

\(256+144-120\,t=0\)

\(400=120\,t\)

\(t=\dfrac {10}{3}\;sec\)

image

A particle is projected with velocity \(u = 20\; m/sec\) making an angle of \(\theta = 37°\) from a height of \(h = 20\; m\), as shown in figure. Calculate the time when final velocity is perpendicular to the initial velocity.

image
A

\(t=\dfrac {10}{3}\;sec\)

.

B

\(t=\dfrac {20}{3}\;sec\)

C

\(t=4\;sec\)

D

\(t=2\;sec\)

Option A is Correct

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