Given, initial speed, u=25ms−1 Angle of projection, θ=45∘
Let, the maximum height =H Time taken to reach, H=t Acceleration due to gravity g=10ms−2 Velocity at maximum height along Y-axis, vy=0ms−1 As we know that, for motion along Y-axis, ∵vy2−uy2=2gH ∴02−(25sin45∘)2=−2×10×H ⇒−312.5=−20H ∵H=15.625m ∴H=