A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it reaches its highest point? (b) What is its velocity 1 s before it reaches its highest point? (c) What is the change in its velocity during this 1-s interval? (d) What is its velocity 1 s after it reaches its highest point? (e) What is the change in velocity during this 1-s interval? (f) What is the change in velocity during the 2-s interval? (Careful!) (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?

Solution 2P Introduction When a body is thrown vertically upward into air from the surface of earth, we can find out the velocity, acceleration, time and distance of the motion by using Newton’s equations of motion. Step 1 The figure 1 shows when a body is thrown vertically upward from the surface of the earth. The arrow mark represents its direction of motion. The figure 2 shows the case when it reaches to its maximum height and figure 3 represents its motion after it reaches to its maximum height. Step 2 a) Velocity of the ball when it reaches to its highest point. When it reaches to its maximum point, the ball is said to be at rest at that particular instant. So, it will stand there at that particular instant hence there is no change in displacement of the body will occur. So, the velocity is “zero t its maximum height. Step 3 b) Velocity of the ball 1 second before it reaches to its highest point. Consider the velocity of the ball as “v” when it reaches to its maximum height at time “t” seconds. Suppose, the velocity of the ball is “v ’ “ when the time is 1 second before the ball reaches to its maximum height that is, “(t-1)” seconds. So, according to Newton’s equations of motion, v = u + at [it is derived from the equation for acceleration, a = (v-u)/t] Where, v - Final velocity u - Initial velocity a - acceleration t - time taken So, the velocity of the ball 1 second before it reaches to its maximum height would be, v ‘ = u - g (t-1) ----------- (2) Since acceleration is acceleration due to gravity here, a = g. We have put a negative sign for g since the motion is against the acceleration due to gravity. Step 4 c) Change in velocity in 1 second interval So, at maximum height, v = u - gt - ---------- (1) Since acceleration is acceleration due to gravity here, a = g. We have put a negative sign for g since the motion is against the acceleration due to gravity. So, the velocity of the ball 1 second before it reaches to its maximum height would be, v ‘ = u - g (t-1) ----------- (2) So, change in velocity will be, v - v ’. That is, v - v ‘ = [u - gt] (m/s) - [u - g (t-1)] (m/s) v - v ‘ = u - gt - u + gt - g = - g m/s That is, v -v ‘ = - g m/s 2 g = 9.8 m/s So, v - v ‘ = - 9.8 m/s