Question

a football kicked in front of goal post at an angle 45°to the ground just clears the top bar of the 3m high.calculate the velocity of projection and the time taken to hit the ground again from the instant it was kicked​

Answer 1

Given :

Angle of projection = 45°

Max. height = 3m

To Find :

Initial velocity of ball and time of flight.

Solution :

❖ A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration is equal to g.

A] Initial velocity of projectile :

It is given that maximum height attained by the ball is 3m.

$\sf:\implies\:H=\dfrac{u^2\sin^2\theta}{2g}$

$\sf:\implies\:3=\dfrac{u^2\sin^2(45^{\circ})}{2(10)}$

$\sf:\implies\:3\times 20=u^2\times \left(\dfrac{1}{\sqrt2}\right)^2$

$\sf:\implies\:u^2=60\times 2$

$\sf:\implies\:u=\sqrt{120}$

$:\implies\:\underline{\boxed{\bf{\orange{u=10.95\:ms^{-1}}}}}$

B] Time of flight :

$\sf:\implies\:T=\dfrac{2u\sin\theta}{g}$

$\sf:\implies\:T=\dfrac{2(10.95)\times\sin(45^{\circ})}{10}$

$\sf:\implies\:T=\dfrac{2.19}{\sqrt2}$

$:\implies\:\underline{\boxed{\bf{\gray{T=1.55\:s}}}}$

Answer 2

Given :

Angle of projection = 45°

Max. height = 3m

To Find :

Initial velocity of ball and time of flight.

Solution :

❖ A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration is equal to g.

A] Initial velocity of projectile :

It is given that maximum height attained by the ball is 3m.

$\sf:\implies\:H=\dfrac{u^2\sin^2\theta}{2g}$

$\sf:\implies\:3=\dfrac{u^2\sin^2(45^{\circ})}{2(10)}$

$\sf:\implies\:3\times 20=u^2\times \left(\dfrac{1}{\sqrt2}\right)^2$

$\sf:\implies\:u^2=60\times 2$

$\sf:\implies\:u=\sqrt{120}$

$:\implies\:\underline{\boxed{\bf{\orange{u=10.95\:ms^{-1}}}}}$

B] Time of flight :

$\sf:\implies\:T=\dfrac{2u\sin\theta}{g}$

$\sf:\implies\:T=\dfrac{2(10.95)\times\sin(45^{\circ})}{10}$

$\sf:\implies\:T=\dfrac{2.19}{\sqrt2}$

$:\implies\:\underline{\boxed{\bf{\gray{T=1.55\:s}}}}$