Question

A projectile is projected with velocity 20 m/s at an angle of 53° with horizontal. Radius of curvature of trajectory of projectile at the instant when its velocity is perpendicular to its initial velocity

Answer 1

u cos 57 = v₀ cos 37°

v₀ = \(\frac{20\times cos53}{cos37^\circ}\) 

v₀ = \(\frac{20\times 3/5}{4/5}\)

v₀ = \(\cfrac{\frac{60}5}{\frac45}\)

v₀ = \(\frac{60\times5}{20}\)

v₀ = 15

R = \(\frac{15^2}{9cos37^\circ}\) , (R = \(\frac{V_0^2}{a_T}\))

a_T = perpendicular component of velocity

g cos 37°

R = \(\cfrac{225}{10\times\frac45}\)

R = \(\frac{225}8\)m