A triangular crate is slided along the inclined face of the tower as shown in the figure. The crate starts with a initial velocity of 10 m/s and slides till B on the inclined surface and it traces a parabolic path to strike the ground at C. If the crate is of 4 kg mass and the height of the tower is 30 m, find the horizontal distance the crate travel's to strike the ground from the starting of its travel. 1) 31.9 m 2) 24.5 m 3) 31.5 m 4) 23.9 m Taking the point A as the datum : TA + VA = TB + VB 1/2(4)(10)B2 = 1/2(4)VB2 - 4(10) VB2 = 140 (m/s)2 VB = 10.95 m/s The horizontal equation of motion from B to C : s= so + vot x = 0 + 10.95 cos30o(t) Also the vertical equation of motion from B to C : s = so + vot + 1/2 act2 0 = 20 - 10.95(1/2)t + 1/2(-9.81)t2 9.81t2 + 10.95t - 40 = 0 t = 1.54 s Substuting in for x we have : x = 10.95 cos30o(t) = 14.6 m The horizontal distance from A to B is 10cos30o and the total horizontal distance travelled is x = 14.6 + 17.32 = 31.9 m So the total horizontal distance travelled from point A to point C is 31.9 m