Question

A projectile thrown with velocity v at an angle....

Answer 1

Question:

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

$\sf{(1) \:v \: sin \: \theta \times 3 }$

$\sf{(2) \: \dfrac{v \: sin \: \theta}{3} }$

$\sf{ (3) \: \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

$\sf{(4) \: \dfrac{v \: sin \: \theta}{ \sqrt{3} }}$

Answer:

Correct option is

$\sf{ (3) \: \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Solution:

• projectile is thrown with a velocity = v
• angle of projection = θ

Let, the maximum height covered by projectile be $\sf{h_m}$

then, Using the formula for maximum height covered by a projectile

$\longmapsto \sf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} } \quad \: eqn(1)$

Now, Let velocity of projectile at a height half of the maximum height covered be

$\sf{v_0}$

then, again using the formula for maximum height

$\longmapsto \sf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

Using equation (1)

$\longmapsto \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\longmapsto \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\longmapsto \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\longmapsto \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\longmapsto \sf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be,

$\longmapsto \sf{v_{(y)}=v_0 \: sin \theta }$

$\longmapsto \sf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

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

$\sf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Hence, Option (3) is correct.