SATHEE: Kinematics 5 Question 5

Kinematics 5 Question 5

7. A body is projected at $t = 0$ with a velocity $10 m s^{- 1}$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t = 1 s$ is $R$ . Neglecting air resistance and taking acceleration due to gravity $g = 10 m s^{- 2}$ , the value of $R$ is

(2019 Main, 11 Jan I)

(a) $10.3 m$

(b) $2.8 m$

(d) $2.5 m$

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Answer:

Correct Answer: 7. (b)

Solution:

Components of velocity at an instant of time $t$ of a body projected at an angle $θ$ is

$v_{x} = u \cos θ + g_{x} t and v_{y} = u \sin θ + g_{y} t$

Here, components of velocity at $t = 1 s$ , is

$\begin{aligned} v_{x} & = u \cos 60^{\circ} + 0 [as g_{x} = 0] \\ = 10 \times \frac{1}{2} = 5 m / s \\ and v_{y} & = u \sin 60^{\circ} + (- 10) \times (1) \\ = 10 \times \frac{\sqrt{3}}{2} + (- 10) \times (1) = 5 \sqrt{3} - 10 \\ \Rightarrow | v_{y} | & = | 10 - 5 \sqrt{3} | m / s \end{aligned}$

Now, angle made by the velocity vector at time of $t = 1 s$

$\begin{aligned} | \tan α | & = | \frac{v_{y}}{v_{x}} | = \frac{| 10 - 5 \sqrt{3} |}{5} \\ \Rightarrow & \tan α = | 2 - \sqrt{3} | or α = 15^{\circ} \end{aligned}$

$∴$ Radius of curvature of the trajectory of the projected body

$\begin{aligned} R = v^{2} / g \cos α = \frac{(5)^{2} + (10 - 5 \sqrt{3})^{2}}{10 \times 0.97} \\ [∵ v^{2} = v_{x}^{2} + v_{y}^{2} and \cos 15^{\circ} = 0.97] \\ \Rightarrow R = 2.77 m \approx 2.8 m \end{aligned}$

Kinematics 5 Question 5

7. A body is projected at $t = 0$ with a velocity $10 m s^{- 1}$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t = 1 s$ is $R$ . Neglecting air resistance and taking acceleration due to gravity $g = 10 m s^{- 2}$ , the value of $R$ is

Answer:

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Kinematics 5 Question 5

7. A body is projected at t=0 with a velocity 10ms−1 at an angle of 60∘ with the horizontal. The radius of curvature of its trajectory at t=1s is R. Neglecting air resistance and taking acceleration due to gravity g=10ms−2, the value of R is

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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7. A body is projected at $t = 0$ with a velocity $10 m s^{- 1}$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t = 1 s$ is $R$ . Neglecting air resistance and taking acceleration due to gravity $g = 10 m s^{- 2}$ , the value of $R$ is