Question

A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height, H of 31.0 m above sea level, directed at an angle θ above the horizontal with an unknown speed v0 A V The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 153.0 m. Assuming that air friction can be neglected, calculate the value of the angle θ Submit Answer Tries 0/99 Calculate the speed at which the rock is launched Submit Answer Tries 0/99 To what height above sea level does the rock rise?

Answer 1

Height = 31 m t = 6 sec Horizontal distance D = 153 m (a) for x-dirn x = u_0 t + 1/2 a t^2 153 = u_0 cos theta (6) u_0 = 25.5/cos theta rightarrow (1) Now, for y-dirn rightarrow - 31 = u_0 sin theta (6) + 1/2 (- 9.8) (6)^2 - 31 = (25.5/cos theta) sin theta (6) - 176.4 tan theta = 0.950 rightdoublearrow theta = tan^-1 (0.950) = 43.53 degree \r\n Now, put the value of theta in equation (1) u_0 = 25.5/cos (43.53 degree) = 35.17 m/s u_0 = 35.17 m/s To find maximum height we have u^2_f = u^2_0 + 2a x 0 = [35.17 sin (43.53 degree)]^2 + 2(- 9.8) x x = (35.17 times 0.6887)^2/2 times 9.8 x = 29.936 m