A projectile is thrown from a point O on the ground at an angle from the vertical and with a speed The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting

Question

A projectile is thrown from a point O on the ground at an angle $45 °$ from the vertical and with a speed $5 \sqrt{2} m/s.$ The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g = 10 {m/s}^{2}$ . The value of $x$ is _________ . [2021]

Smart Achievers · Accepted Answer

(7.5)

$Range R = \frac{2 u_{x} u_{y}}{g} = \frac{2 \times 5 \times 5}{10} = 5 m$

$Time of flight T = \frac{2 u_{y}}{g} = \frac{2 \times 5}{10} = 1 s$

$∵$ $Time of motion of one part falling vertically downwards = 0.5 s = \frac{T}{2}$

$From law of conservation of momentum, P_{i} = P_{f}$

$2 m \times 5 = m \times v \Rightarrow v = 10 m/s$

As there is no initial velocity in vertical direction for second particle, so it will also take same time as first one.

$∴$ $Time of motion of another part, t = \frac{T}{2} = 0.5 s$

$Displacement of other part in 0.5 s in horizontal direction = v \frac{T}{2} = 10 \times 0.5 = 5 m = R$

$∴$ $Total distance of second part from point ‘O’ is,$

$x = \frac{3 R}{2} = 3 \times \frac{5}{2} \Rightarrow x = 7.5 m$

Solution

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