A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m/s at an angle of 33.0o above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw x-t, y-t, vx-t, and vy-t graphs for the motion.

Solution 21 E Step 1 : In this question, we need to find the Maximum height reached by the stone y 0 The speed of rock vertically v y Horizontal range from base of the building to stone x 0 Let us consider the data given Height of the building y = 15 m Initial speed of the rock v =i30 m/s Angle above horizontal = 33 0 Step 2 : Let us consider graphical representation of the situation Step 3: Finding the x-component of the initial speed This is given by v = v cos xi i Substituting the values we get 0 vxi 30 m/s × cos 33 vxi 30 m/s × 0.8367 vxi 25.160 m/s Similarly finding the y component This is given by vyi v sii Substituting the values we get v = 30 m/s × sin 33 0 yi vyi 30 m/s × 0.5446 v = 16.33 m/s yi Step 4 : Let us find the maximum distance reached by the rock above the horizontal It is obtained using y = v sin ()/2g 0 0 Here y 0 maximum height along y axis v initial velocity i 2 g = gravitational force = 9.8 m/s = angle along the horizontal The above equation can also be written as 2 (vyi y 0 2g Substituting the values we get 2 (16.33 m/s ) y0= 2.×9.8 m/s 2 2 y = 266.6689 2 /s 0 19.6 m/s y0= 13.60 m