Question

Part A The water sprinkler, positioned at the base of a hill, releases a stream of water with a velocity of v-35 ft/s as shown in (Figure 1). Assume that the hill is defined by the equation y-(0.05a*) ft and neglect the size of the sprinkler. Determine the coordinates of the point B where the water strikes the ground on the hill. Enter the and y coordinates of the point separated by a comma using three significant figures. ft Submit Request Answer

Answer 1

The first derivative of the horizontal component of the velocity water sprinkler v_x = v cos 60 degree = 35 cod 60 degree v_x = 17.5 ft/s The first derivative of the vertical component of the velocity water sprinkler v_y = v sin 60 degree = 35 sin 60 degree v_y = 30.31 ft/s The range of the water sprinkler was positioned at the base of hill in the vertical motion equation s = v_xt x = 17.5 t t = x/17.5 doublerightarrow 0.0571 x \r\n The range of the water sprinkler which was positioned at the base of the hill in the vertical motion equation s = s_0 + v_0t + 1/2 a_ct^2 y = 0 + 30.31t + 1/2 (-32.2)t^2 doublerightarrow 0.05 x^2 = 0 + 30.31 (0.0571x) + 1/2 (-32.2)(0.0571x)^2 doublerightarrow 0.05x^2 = 1.7307x - 0.0525x^2 doublerightarrow 0.1025 x^2 - 1.7307x = 0 doublerightarrow (0.1025x - 1.7307)x = 0 x = 16.884 ft y = 0.05x^2 = 0.05(16.884)^2 doublerightarrow 14.253 ft