Question

A person on a cliff throws a ball with an initial velocity of 30.2 m/s at an angle of 46.50 above the horizontal. If the initial height of the ball, compared to it impact location is, Y = 18.5 m, what is the maximum height it obtains, and where does it land? (ind both as measured from the base of the cliff Max height-Number Units Distance, X NumberUnits

Answer 1

Max height where vertical velocity is zero V_y = 0 U_x = 30.2 times cos (46.5 degree) = 20.788 m/s U_y = 30.2 times sin (46.5) = 21.906 m/s at max height V_y = 0 U_y = 21.906 m/s a = -9.8 m/s^2 V^2 = U^2 - 2 a s 0 = (21.906)^2 - 2 times 9.8 times h h = 24.483 m h_total = 18.5 + 24.483 m H implies 42.983 m (ii) x = ? R = U_x times t implies 20.788 t at the landing point, \r\n displacement in y-direction: - y = -18.5 mater s = Ut + 1/2 at^2 - 18.5 m = (21.906) t - 1/2 times 9.8 t^2 4.9 t^2 - 21.906 t - 18.5 = 0 t = 5.197 second R = U_x times t = (20.788 times 5.197) = 108.03 m x = 108.03 m