Question

An airplane has an average acceleration af 15 m/s from rest, how much time doest take to reach a speed of 325%. sduring takeoff. T the air plane starts A golf ball is kicked from a 8m hill, at an angle of GO。witn thu honzonal with an initiol Sp eed of 25/s a) what''s the intal vertica/ speed ny nad vetical d ight boe tu ground tHow long doeo it take to it he ground?

Answer 1

Given that initial velocity of airplane u = 0 m/s final velocity of airplane v = 325 m/s acceleration a = 15 m/s^2 we know that a = v - u/t Rightarrow t = v - u/a Rightarrow t = 325 - 0/15 Rightarrow t = 65/3 = 21.67 sec therefore t = 21.675 given initial velocity u = 25 m/s theta = 60 degree with x-axis a) initial vertical speed = u sin theta = 25 sin 60 = 25 squareroot 3/2 m/s u_y = 25 squareroot 3/2 m/s = 21.65 m/s \r\n b) s = -8m a = -g = -9.8 m/s^2 u_y = 25 squareroot 3/2 m/s v_y = ? v_y^2 - u^2_y = 2as Rightarrow v^2_y - 25^2 times 3/4 = 2 times 9.8 times 8 Rightarrow v^2_y = 2502.2/4 Rightarrow v_y = squareroot 2502.2/4 = -25.03 m/s -ve indicates download direction therefore v_y = -25.03 m/s