Question

A body is thrown from a point with speed 50 m/s at an angle 37º with horizontal. When it has moved a horizontal distance of 80 m then its distance from point of projection is –

• A. 40 m
• B. 40 $\sqrt{2}$m
• C. 40 $\sqrt{5}$m
• D. None

Answer 1

Answer: (C)

40 $\sqrt{5}$m

Answer 2

As y = x tan q –$\frac{gx^{2}}{2u^{2}\cos^{2}\theta}$

y = 80 ×$\frac{3}{4}$–$\frac{10 \times 80 \times 80 \times 25}{2 \times 50 \times 50 \times 16}$= 40 m

\ distance from point of projection

=$\sqrt{(80)^{2} + (40)^{2}}$m