Question

A body is projected from ground with a speed of 40 m/s making an angle of 30^o with horizontal. The body crosses a horizontal line twice in a time interval $Δ t$ . If the height of the line above the ground is 15 m, find $Δ t$ . [Take g = 10 m/s²]

Moderate

Solution

The body crosses the horizontal line PQ with velocity v at A and B
$∴ {(V \sin θ)}^{2} = {(u \sin α)}^{2} - 2. g . h = {(40 \sin 30^{o})}^{2} - 2.10.15$
$\begin{array}{l} \Rightarrow V \sin θ = 10 m / s \\ ∴ Δ t = \frac{2 V \sin θ}{g} = \frac{2 \times 10}{10} \sec = 2 \sec \end{array}$

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