1.

The speed of a projectile (u) rekuces by ` 50 %` on reachig maximum hight. What is the range on the horizontal plane ?

Answer» If ` theta` is the angle of projection, then velocity of projectile at height point `= u cos theta`
As per question, ` u cos theta = ( 50)/( 100) u = 1/ 2 u ` or ` cos theta = 1/2 = cos 60^@` or ` theta =60^@`
Horizontal range, `R= ( u^2 sin 2 theta)/g = u^2 /g sin 2 xx 60^@ = u^2 /g sin 120^@ = u^2/g xx sin ( 180^@- 60^@)`
` u^2/g sin 60^@ = u^@/g 60^2 = u^2/g xx (sqrt 3)/2`.


Discussion

No Comment Found

Related InterviewSolutions