
As ball is projected at an angle 45∘ to the horizontal therefore Range =4H or 10=4H⇒H=410=2.5 m (∵ Range =4 m+6 m=10 m) Maximum height, H=2 gu2sin2θ ∴u2=sin2θH×2 g=(21)22.5×2×10=100 or, u=100=10 ms−1 Height of wall PA =OAtanθ−21u2cos2θg(OA)2=4−21×10×10×n1×0110×16=2.4 m