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Standard XII

Physics

The Equation for the Path of Projectile

A particle is...

Question

A particle is projected from ground with some initial velocity making an angle of $45^{o}$ with the horizontal. It reaches a height of $7.5 m$ above the ground while it travels a horizontal distance of $10 m$ from the point of projection. The initial speed of the projection is

5 m / s

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10 m / s

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20 m / s

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40 m / s

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Solution

The correct option is C $20 m / s$
Given that,

Angle $θ = 45^{0}$

Horizontal distance $x = 10 m$

Vertically distance $y = 7.5 m$

Acceleration due to gravity $g = 10 m / s^{2}$

Now,

$y = (tan θ) x - (\frac{g}{2 u^{2} {cos}^{2} θ}) x^{2}$

$7.5 = 1 \times 10 - \frac{10}{(2 u^{2} \times \frac{1}{2})} \times 100$

$7.5 - 10 = - \frac{1000}{u^{2}}$

$u^{2} = \frac{1000}{2.5}$

$u = 20 m / s$

Hence, the initial speed of the projection is $20 m / s$

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A particle is projected from ground with some initial velocity making an angle of 45o with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. The initial speed of the projection is

The correct option is C 20 m/sGiven that, Angle θ=450 Horizontal distance x=10m Vertically distance y=7.5m Acceleration due to gravity g=10m/s2 Now, y=(tanθ)x−(g2u2cos2θ)x2 7.5=1×10−10(2u2×12)×100 7.5−10=−1000u2 u2=10002.5 u=20m/s Hence, the initial speed of the projection is 20 m/s

A particle is projected from ground with some initial velocity making an angle of $45^{o}$ with the horizontal. It reaches a height of $7.5 m$ above the ground while it travels a horizontal distance of $10 m$ from the point of projection. The initial speed of the projection is