A body is projected 3 m from a wall of height 5 m- If the ball has to just clear the wall find - The ball hits the ground at a distance of 12 m from the wall A- -tan -125-12B- -tan -126-12C- -tan -122-12D- -tan -121-12

The correct option is A R = 3 + 12 = 15 m Now, y = x tan θ [ 1 x/R ] 5 = 3 tan θ [ 1 3/15 ] ⇒θ = tan^1 [ 25/12 ] ...

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Standard XII

Physics

Work Energy Theorem Application

A body is pro...

Question

A body is projected 3 m from a wall of height 5m. If the ball has to just clear the wall find θ. The ball hits the ground at a distance of 12 m from the wall

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Solution

The correct option is A
R = 3 + 12 = 15 m
$N o w, y = x t a n θ [1 - \frac{x}{R}] 5 = 3 t a n θ [1 - \frac{3}{15}] \Rightarrow θ = t a n^{1} [\frac{25}{12}]$

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A body is projected 3 m from a wall of height 5m. If the ball has to just clear the wall find θ. The ball hits the ground at a distance of 12 m from the wall

The correct option is A R = 3 + 12 = 15 m Now, y=xtan θ[1−xR]5=3 tan θ[1−315]⇒θ=tan1[2512]

The correct option is A
R = 3 + 12 = 15 m
$N o w, y = x t a n θ [1 - \frac{x}{R}] 5 = 3 t a n θ [1 - \frac{3}{15}] \Rightarrow θ = t a n^{1} [\frac{25}{12}]$