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Question
A player throws a ball upwards with an initial speed of 29.4 m s–1.
- What is the direction of acceleration during the upward motion of the ball?
- What are the velocity and acceleration of the ball at the highest point of its motion?
- Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
- To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
Solution
- Since gravity is at work while the ball is in motion, the direction of gravity-related acceleration is always vertically downward.
- At maximum height, the velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value, i.e., 9.8 m/s2.
- If we consider highest point of ball motion as x = 0, t = 0 and vertically downward direction as the positive direction of x-axis, then
- When moving upward, the position sign is negative, the velocity sign is negative, and the acceleration sign is positive, which means that v < 0 and n > 0.
- When moving downward, the position sign is positive, the velocity sign is positive, and the acceleration sign is also positive, which means that v > 0 and a > 0.
-
Initial velocity of the ball, u = 29.4 m/s
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)
Acceleration, a = – g = – 9.8 m/s2
From third equation of motion, height (s) can be calculated as:
`v^2- u^2 = 2gs`
`s = (v^2 - u^2)/2g`
`=((0)^2 - (29.4)^2)/(2xx(-9.8)) = 44.1 m`
From first equation of motion, time of ascent (t) is given as:
v = u + at
`t = (v-u)/a = (-29.4)/-9.8 = 3 s`
Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s
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