A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____?

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12-12-2021
Physics

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____?

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ThαtButtєřflíєs ThαtButtєřflíєs · Answer 1 · 2021-12-12T13:27:10+01:00

Let, the maximum height covered by projectile be [tex]\sf{H_m}[/tex]

[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be [tex] \sf{v_0}[/tex]

[tex]\qquad[/tex]______________________________

Then –

[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]

Now, the vertical component of velocity of projectile at the height half of [tex] \sf{h_m}[/tex] will be –

[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]

Therefore, the vertical component of velocity of projectile at this height will be–

☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]

jodishakg jodishakg · Answer 2 · 2021-12-12T13:36:07+01:00

jodishakg jodishakg

12-12-2021

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____? ​

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____?