Question

2. A shell as fired at an angle of 30° to the horizontal
with a velocidy of 19.6 ms find its (1) Horizontal range
(2) Maximum height (given g=9.8 ms )​

Answer 1

Given :

• Angle of Projection (θ) = 30°.

• Velocity of the shell (u) = 19.6 m/s.

• Acceleration due to gravity (g) = 9.8 m/s².

To find :

• Horizontal Range (R).

• Maximum height (h).

Solution :

To find the horizontal range of the shell :

We know the Equation for Horizontal range i.e,

$\boxed{\bf{R = \dfrac{u^{2}\sin2\theta}{2g}}}$

Where :

• R = Range

• u = Initial Velocity

• θ = Angle of Projection

• g = Acceleration due to gravity

Now using the equation for horizontal range and substituting the values in it, we get :

$:\implies \bf{R = \dfrac{u^{2}\sin2\theta}{2g}}$

$:\implies \bf{R = \dfrac{19.6^{2} \times \sin2\theta}{2 \times 9.8}}$

$:\implies \bf{R = \dfrac{19.6 \times 19.6 \times \sin2\theta}{19.6}}$

$:\implies \bf{R = 19.6 \times \sin2\theta}$

We know that ,

$\bf{\sin2\theta = 2sin\theta cos\theta}$

By substituting it in the equation , we get :

$:\implies \bf{R = 19.6 \times 2sin\theta cos\theta}$

$:\implies \bf{R = 19.6 \times 2sin30^{\circ}cos30^{\circ}}$

$:\implies \bf{R = 19.6 \times 2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}}\quad \bf{\bigg[\because \bigg(sin30^{\circ} = \dfrac{1}{2}\bigg)\:;\:\bigg(cos30^{\circ} = \dfrac{\sqrt{3}}{2}\bigg)\bigg]}$

$:\implies \bf{R = 19.6 \times \not{2} \times \dfrac{1}{\not{2}} \times \dfrac{\sqrt{3}}{2}}$

$:\implies \bf{R = 19.6 \times \dfrac{\sqrt{3}}{2}}$

$:\implies \bf{R = 9.8 \times \sqrt{3}}$

$:\implies \bf{R = 9.8\sqrt{3}}$

$\boxed{\therefore \bf{R = 9.8\sqrt{3}\:m}}$

Hence the Horizontal Range of the shell is 9.8√3 m.

To find the Maximum height reached :

We know the third Equation of Motion ,i.e,

$\boxed{\bf{v^{2} = u^{2} \pm 2gh}}$

Where :

• v = Final Velocity.
• u = Initial Velocity.
• g = Acceleration due to gravity.
• h = Height.

From the third Equation of Motion , we get :

$:\implies \bf{v^{2} = u^{2} \pm 2gh}$

[Note : Since , the object is moving upward ,i.e, against the gravity, the acceleration due to gravity will be taken as negative]

$:\implies \bf{v^{2} = u^{2} - 2gh}$

We know that,

$\bf{v = u cos\theta}$

So , substituting it in the equation , we get :

$:\implies \bf{(u cos\theta)^{2} = u^{2} - 2gh}$

$:\implies \bf{u^{2}cos^{2}\theta = u^{2} - 2gh}$

$:\implies \bf{0 = u^{2} - u^{2}cos^{2}\theta - 2gh}$

$:\implies \bf{0 = u^{2}(1 - cos^{2}\theta) - 2gh}$

We know that ,

$\bf{1 - cos^{2}\theta = sin^{2}\theta}$

So , substituting it in the equation , we get :

$:\implies \bf{0 = u^{2}sin^{2}\theta - 2gh}$

$:\implies \bf{- u^{2}sin^{2}\theta = - 2gh}$

$:\implies \bf{\not{-} u^{2}sin^{2}\theta = \not{-} 2gh}$

$:\implies \bf{u^{2}sin^{2}\theta = 2gh}$

$:\implies \bf{\dfrac{u^{2}sin^{2}\theta}{2g} = h}$

$:\implies \bf{h_{max} = \dfrac{u^{2}sin^{2}\theta}{2g}}$

Now by substituting the values in it, we get :

$:\implies \bf{h_{max} = \dfrac{19.6^{2} \times sin^{2}30^{\circ}}{2 \times 9.8}}$

$:\implies \bf{h_{max} = \dfrac{19.6 \times 19.6 \times sin^{2}30^{\circ}}{19.6}}$

$:\implies \bf{h_{max} = 19.6 \times sin^{2}30^{\circ}}$

$:\implies \bf{h_{max} = 19.6 \times \bigg(\dfrac{1}{2}\bigg)^{2}} \quad \bf{\bigg(\because sin30^{\circ} = \dfrac{1}{2}}\bigg)$

$:\implies \bf{h_{max} = 19.6 \times \dfrac{1}{4}}$

$:\implies \bf{h_{max} = 4.9}$

$\boxed{\therefore \bf{h_{max} = 4.9\:m}}$

Hence the maximum height reached is 4.9 m.