Question

A motocross hike drives off of a jump angled at 45 degree. When the bike leaves the jump, it has a speed of 20 m/n (about 40 mph). To ensure a smooth landing, the landing ramp nerds to be at the same angle as the hikes trajectory when it lands. If the landing ramp is 15 meters lower than the jump, what angle below the horizon should the Undoing ramp slope down? -30 degree -12 degree -58 degree -45 degree Astronaut Mark Tawney is very sick of potatoes Therefore, he throws a potato at a 30 degree angle above the horizontal as hard as he can. The initial height of the potato when it leaves his hand is 2.0 meters. and its initial velocity is 5.0 m/s (about 10 mph). Gravitational acceleration on the surface of Mars is 3.8 m/s^2. How far will wanted have to walk when he goes to retrieve the potato so he doesn't starve? (Assume that the ground wanted is standing on is flat and Level.) 9.5 m 5.6 m 10.4 m 8.1 m
Please answer both

Answer 1

From the kinematic equation V^2 - u^2 = 2as V^2_y = u^2 + 2as = (V sin theta)^2 + 2as = squareroot (10 squareroot 2)^2 + (2 times 9.3 times 15)^2 V_y = 22.23 m/sec horizontal component of the velocity is V_x = V cos theta = 20 cos (45) = 14.142 m/sec Now find the angle is theta = tan^- 1 (V_y /V_x) = tan^- 1 (22.23/14.142) = tan^- 1(1.5719) = 57.53 approximately 58 degree below the origin so angle = - 58 degree option (c) correct \r\n time to reach A to b is time of the flight is t_1 = 2v sin theta /g = 2 times 5 times sin (30)/3.8 = 1.316 sec Time to reach B to C from kinematic equation s = ut + 1/2 at^2 qm = (5m/s middot sin(30)) t_2 - 1/2 (3.8) (t^2_2) 1.9 t^2_2 + 2.5 t_2 - 2 = 0 solving sor t by using quadratic equation t = 0.561 sec total time of flight (T) = t_1 + T_2 = 1.316 - 0.561 = 1.877 sec