Question

27. A rock is thrown with a velocity of 10 m/s from the edge of a cliff that is a height of 40 m above level ground. Neglect air resistance. In order for the rock to travel a maximum horizontal distance from the cliff before hitting the ground, what angle should the rock be thrown?

Answer 1

Using Delta y = u_0 y t + 1/2 gt^2 therefore -h = u_0 sin theta t - 1/2 g times t^2 equation (i) Now R = u_0 X t R/u_0 cos theta = t. therefore from equation 1 -h = u_0 sin theta/u_0 cos theta R - 1/2 g(R^2/u_0^2 cos^2 theta) -h = tan theta R - 1/2 g R^2/u_0^2 cos^2 theta -2h cos^2 theta = 2 cos theta tan theta R - gR^2/u_0^2 or sin plying h(1 + cos 2 theta) = gR^2/u_0^2 - R sin 2 theta. Now, dR/d theta = 0 so, h tan 2 theta = R implies tan 2 theta = R/h implies theta = 1/2 tan^-1 (R/h) therefore h = gR^2/2u_0^2 - u_0^2/2g implies R = squareroot 2u_0^2/g (h + u_0^2/2g) therefore R_max = squareroot 2 times 10^2/9.81 (40 + 10^2/2 times 9.81) implies R_max = 30.32 m theta_max = 1/2 tan^-1 (30.32/40) theta = 18.58 degree