1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A particle projected from the level ground just cleared in its Ascent a wall 30 M high and 120√3 m away measured horizontally. the time Since projection to clear the wall is 2 second It will strike the ground in the same horizontal plane from the wall on the other side at a distance of 150√3 120√3 180√3 210√3

Open in App
Solution

## Let the velocity of particle is u and projected with angle θ now for horizontal equation distance = speed x time 120√3 = ucosθ​ x 2 ucosθ = 60√3 -------(1) now for vertical motion 30 = usinθ ×2 −½×10×2² 30 = 2usinθ −20 2usinθ =50 2usinθ/g = 50/10 = 5 --------(2) now the formula of range is R = 2usinθ×ucosθ/g R = 5 ×60√3 (putting values from eq 1,2) R =300√3 metre R = 519.60 metre now the distance of other side from the wall is distance = 519.60 -120 x 1.732 = 519.60 - 207.84 =311.76 metre =180√3 So the answer is 180√3

Suggest Corrections
5
Join BYJU'S Learning Program
Related Videos
Motion Under Constant Acceleration
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program