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Standard XII

Physics

Summary for Time, Height and Range

A body is thr...

Question

A body is thrown from a point with speed $50 m/s$ at an angle $37^{\circ}$ with horizontal. When it has moved a horizontal distance of $80 m$ then its distance from point of projection is

40 m

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40 \sqrt{2} m

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40 \sqrt{5} m

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Solution

The correct option is C $40 \sqrt{5} m$
As we know that the equation of trajectory is $y = x tan θ - \frac{g x^{2}}{2 u^{2} {cos}^{2} θ}$ , where $y$ is the verticle displacement, $x$ is horizontal displacement.
Putting the values in the equation of trajectory we get,

$y = 80 \times \frac{3}{4} - \frac{10 \times 80 \times 80 \times 25}{2 \times 50 \times 50 \times 16} = 40 m$
$(tan 37^{o} = \frac{3}{4}, u = 50 m/s, x = 80 m$ and $cos 37^{o} = \frac{4}{5})$

$∴$ Distance from point of projection
$= \sqrt{(80)^{2} + (40)^{2}} m = 40 \sqrt{5} m$

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A body is thrown from a point with speed 50 m/s at an angle 37∘ with horizontal. When it has moved a horizontal distance of 80 m then its distance from point of projection is

A body is thrown from a point with speed $50 m/s$ at an angle $37^{\circ}$ with horizontal. When it has moved a horizontal distance of $80 m$ then its distance from point of projection is