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The projectile is fired with initial velocity of 100ms-1 at an angle of 30° with the horinzontal.Calculate a.The time of flight b.The maximum height attained c.The range​

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Starrysoul100

[tex]\bold{\huge{\underline{ Solution }}}[/tex]

Here,

  • The projectile is fired with initial velocity =100m/s
  • The Angle formed with the horizontal = 30°

Answer (a)

Here, we have

  • Initial velocity = 100 m/s
  • Angle of projection = 30°

We have to find the time of flight

We know that,

Time of flight

[tex]\sf{ = }{\sf{\dfrac{ 2uSin{\theta}}{g}}}[/tex]

Subsitute the required values,

[tex]\sf{ = }{\sf{\dfrac{ 2{\times}100{\times}Sin30{\degree}}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 200{\times}{\frac{1}{2}}}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 100}{9.8}}}[/tex]

[tex]\bold{ = 10.20 \: s }[/tex]

Hence, The time of flight is 10.20 sec .

Answer (b)

Here, We have

  • Initial velocity = 100 m/s
  • Angle of projection = 30°

We have to find the maximum height attained by the body

We know that,

Maximum height

[tex]\sf{ = }{\sf{\dfrac{ u^{2}Sin^{2}{\theta}}{2g}}}[/tex]

Subsitute the required values,

[tex]\sf{ = }{\sf{\dfrac{ (100)^{2}{\times}Sin^{2}30{\degree}}{2{\times} 9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 10000{\times}{\frac{1}{2}}{\times}{\frac{1}{2}}}{19.6}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 5000{\times}{\frac{1}{2}}}{19.6}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 2500}{19.6}}}[/tex]

[tex]\sf{ = 127.5 \: m}[/tex]

Hence, The maximum height attained by the body is 127.5 .

Answer ( c) :-

Here, we have

  • Initial velocity = 100 m/s
  • Angle of projection = 30°

We have to find the horizontal range

We know that,

Horizontal range

[tex]\sf{ = }{\sf{\dfrac{ u^{2}Sin2{\theta}}{g}}}[/tex]

Subsitute the required values,

[tex]\sf{ = }{\sf{\dfrac{ (100)^{2}{\times}Sin2{\times}30{\degree}}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 10000{\times}Sin{\times}60{\degree}}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 10000{\times}{\frac{\sqrt{3}}{2}}}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 5000{\times}\sqrt{3}}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 5000{\times}1.732}{9.8}}}[/tex]

[tex]\sf{ = }{\sf{\dfrac{ 8660}{9.8}}}[/tex]

[tex]\sf{ = 883.67 \: or 883.7 m}[/tex]

Hence, The range of the body is 883.67 or 883.7 m.

semsee45

Answer:

(a) 10.20 s (nearest hundredth)

(b) 127.55 m (nearest hundredth)

(c) 883.70 m (nearest hundredth)

Explanation:

Part (a)

At the end of the projectile's flight, its vertical displacement will be zero.

Resolving vertically, taking up as positive:

[tex]s=0\quad u=100 \sin30^{\circ} \quad v=v, \quad a=-9.8, \quad t=t[/tex]

[tex]\begin{aligned}\textsf{Using }\:s & =ut+\dfrac12at^2:\\ 0 & =100 \sin 30^{\circ}t+\dfrac12(-9.8)t^2\\ 0 & = 50t-4.9t^2\\ 4.9t^2 & = 50t\\ 4.9t & = 50\\ t & = \dfrac{50}{4.9}\\ t & = 10.20\:\sf s\:(nearest\:hundredth)\end{aligned}[/tex]

Part (b)

At the maximum height, vertical velocity will be zero.

Resolving vertically, taking up as positive:

[tex]s=0\quad u=100 \sin30^{\circ} \quad v=0, \quad a=-9.8, \quad t=t[/tex]

[tex]\begin{aligned}\textsf{Using }\:v^2 & = u^2+2as :\\ 0^2 & = (100 \sin 30^{\circ})^2+2(-9.8)s\\ 0 & = 2500-19.6s\\ 19.6s & = 2500\\ s & = \dfrac{2500}{19.6}\\ s & = 127.55\: \sf m\:(nearest\:hundredth) \end{aligned}[/tex]

Part (c)

The horizontal velocity of a projectile is always constant, so u = v.

The horizontal component of acceleration is zero.

Resolving horizontally, taking right as positive (and using the value for t we found in part a):

[tex]s=s\quad u=100 \cos30^{\circ} \quad v=100 \cos30^{\circ} , \quad a=0, \quad t=\dfrac{50}{4.9}[/tex]

[tex]\begin{aligned}\textsf{Using }\:s & =ut+\dfrac12at^2 : \\ s & =(100 \cos 30^{\circ})\left(\dfrac{50}{4.9}\right)+\dfrac12(0)\left(\dfrac{50}{4.9}\right)^2\\ s & =50\sqrt{3}\left(\dfrac{50}{4.9}\right)+0\\ s & =883.70\: \sf m\:(nearest\:hundredth)\end{aligned}[/tex]

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