A particle is thrown with initial speed u at angle ‘θ’, with the

1.	A particle is thrown with initial speed u at angle ‘θ’, with the horizontal, under gravity. The relation between maximum height ‘h’ and range upon level ground ‘R’ is
Answer» Explanation: Correct option is D 2 u 1+3cos 2 θ The maximum HEIGHT of a projectile, H= 2g u 2 sin 2 θ . . . . . . . . .(1) Range of a projectile, R= 2g 2u 2 sinθ.cosθ . . . . . . . . . .(2) By Pythagoras theorem in ΔOAB d= H 2 + 4 R 2 . . . . . . .(3) Time to react the maximum height A t= 2vcosθ R . . . . . . . . .(4) Average velocity of the particle between its POINT of projection and highest point of TRAJECTORY is v avg = t d =2vcosθ 4 1 +( R H ) 2 .. . . . . . .(5) Divide equation (1) by (2) R H = 4 tanθ . . . . . . . . .(6) Substitute equation (6) in equation (5), we get v avg = 2 u 1+3cos 2 θ The correct option is D.

A particle is thrown with initial speed u at angle ‘θ’, with the horizontal, under gravity. The relation between maximum height ‘h’ and range upon level ground ‘R’ is

Answer»

Explanation:

Correct option is

1+3cos

The maximum HEIGHT of a projectile,

sin

. . . . . . . . .(1)

Range of a projectile,

sinθ.cosθ

. . . . . . . . . .(2)

By Pythagoras theorem in ΔOAB

. . . . . . .(3)

Time to react the maximum height A

2vcosθ

. . . . . . . . .(4)

Average velocity of the particle between its POINT of projection and highest point of TRAJECTORY is

avg

=2vcosθ

)

.. . . . . . .(5)

Divide equation (1) by (2)

tanθ

. . . . . . . . .(6)

Substitute equation (6) in equation (5), we get

avg

1+3cos

The correct option is D.