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

Answer 1

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