If u is the initial velocity- of a body projected with an angle - with a horizontal- then the maximum height breached is -

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Application of Projectile to a Height

If u is the i...

Question

If u is the initial velocity, of a body projected with an angle $θ$ with a horizontal, then the maximum height breached is ____ .

\frac{u^{2}}{2}

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\frac{u^{2} s i n θ}{2 g}

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\frac{u s i n θ}{2 g}

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\frac{u^{2} {s i n}^{2} θ}{2 g}

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Solution

The correct option is D $\frac{u^{2} {s i n}^{2} θ}{2 g}$
Taking upward direction to be positive.
Vertical direction :
Initial velocity $u_{y} = u sin θ$
Acceleration $a_{y} = - g$
At maximum height $H$ , final vertical velocity is zero i.e. $v_{y} = 0$
Using $v_{y}^{2} - u_{y}^{2} = 2 a S$
$∴$ $0 - u^{2} {sin}^{2} θ = 2 (- g) H$
$⟹ H = \frac{u^{2} {sin}^{2} θ}{2 g}$

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If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ :

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ . [ $g$ is the acceleration due to gravity]

Q. Assertion :When a body is projected at an angle

45^{0}

, its maximum height is half of horizontal range. Reason: Horizontal range

= \frac{U^{2} s i n 2 θ}{g}

and maximum height

= \frac{U^{2} s i n^{2} θ}{2 g}

Q. If a body is thrown up with an initial velocity u and covers a height of h, then

h = \frac{u^{2}}{2 g}

Q. Assertion :Maximum possible height attained by a projectile is

\frac{u^{2}}{2 g}

, where u is initial velocity Reason: To attain maximum height, a particle is thrown vertically upwards at

θ = 90^{o}

to the ground.

H = \frac{u^{2}}{2 g} s i n^{2} θ \Rightarrow H_{m a x} = \frac{u^{2}}{2 g}

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If u is the initial velocity, of a body projected with an angle θ with a horizontal, then the maximum height breached is ____ .

The correct option is D u2sin2θ2gTaking upward direction to be positive.Vertical direction :Initial velocity uy=usinθAcceleration ay=−gAt maximum height H, final vertical velocity is zero i.e. vy=0Using v2y−u2y=2aS∴ 0−u2sin2θ=2(−g)H⟹ H=u2sin2θ2g

If u is the initial velocity, of a body projected with an angle $θ$ with a horizontal, then the maximum height breached is ____ .