An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 s after it leaves the thrower’s hand. Ignore air resistance. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch ay-t, vy-t, and y-t graphs for the motion of the egg.
Solution 46E Let us have a look at the following figure. Suppose the time taken by the egg to reach the top height h is t and the time back to the point of start is also the same. Since the time of flight is 5.00 s, hence the time taken to reach a point below 30 m of the origin is (5-t-t) s or (5-2t) s. Now, at the maximum height, initial speed u = gt…..(1) , since final speed is zero. Now, when the egg drops to the point of origin from the maximum height, 1 2 h = 2t …..(2) When the egg falls to 30 m below the point of origin, 1 2 h + 30 = g25 2t) …..(3) Substituting the value of h from equation 2 in equation 3, 1gt + 30 = (25 + 4t 20t) 2 2 2 Using g = 9.8 m/s and solving for t, t = 0.805 s (a) At the maximum height, final speed v = 0 From the equation, v = u gt u = gt u = 9.8 × 0.805 s u = 7.9 m/s6 (b) We can calculate the height from equation (2), h = gt 2 2 h = 1 × 9.8 × (0.805) m 2 h = 3.17 m This is the height of the egg above the starting point. (c) he magnitude of velocity at the highest point is zero. 2 (d) The magnitude of acceleration at the highest point is 9.8 m/s and it will act downward. (e) e can show the following plots for the ay-t, vy-t, and y-t graphs