Question

A projectile is thrown with a velocity of 10 ms⁻¹ at an angle of 30° with horizontal. The value of maximum height gained by it is *
(a) 1 m
(b) 1.25 m
(c) 2 m
(d) 2.5 m​

Answer 1

Given:

$\sf{Initial \:Velocity=10m}{s^-1}$

Angle of Projection=30°

To Find:

$\sf{Maximum \:Height \:Gained \:by \:the \:Object.}$

Solution:

$\sf{Maximum \:Height=}{\dfrac{{u^2}{sin^2\theta}}{2g}}$

$\sf{Maximum \:Height=}{\dfrac{100\times0.25}{19.6}}$

$\sf{Maximum \:Height=}{\dfrac{25}{19.6}}$

$\sf{Maximum \:Height=1.275m}$

$\sf\underline{Thus, \:maximum \:height \:attained \:by \:the \:body \:is \:1.275m}$

Answer 2

Answer:

Given:

Initial velocity (u) = 10m/s.

Angle of projection = 30°.

To Find:

Maximum height of projectile.

Solution:

We know that,

$\star \: \boxed{\rm\red{Height \: of \: projectile = \frac{ {u}^{2} {sin \theta}^{2} }{2g}}}$

Substituting values,

$:\implies\mathtt{ \frac{( {10)}^{2} \times ( {sin30)}^{2} }{2 \times 9.8} }$

$:\implies\mathtt{ \frac{100 \times 0.5 \times 0.5}{19.6} }$

$:\implies\mathtt{ \frac{100 \times 0.25}{19.6} }$

$:\implies\mathtt{\dfrac{\cancel{25}}{\cancel{19.6}}}$

$\sf\star \: \underbrace\green{Height \: of \: projectile \: = \: 1.275m} \: \star$

Hence,

Value of maximum height gained is 1.275m.

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