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A ball is thrown vertically upwards with a speed of 19.6 m/s from the top of a tower and returns to the earth in 6 seconds. Find the height of the tower (g = 9.8 m/s²).

a. 19.6 m
b. 38.2 m
c. 58.8 m
d. 78.4 m

Answer :

Final answer:

The height of the tower from which a ball thrown vertically upwards returns to Earth in 6 seconds is calculated using kinematic equations, resulting in a height of d) 78.4 meters.

Explanation:

To find the height of the tower from which a ball thrown vertically upwards with a speed of 19.6 m/s returns to Earth in 6 seconds, we utilize kinematic equations for uniformly accelerated motion.

The total time taken for the ball to go up and then come back down is 6 seconds. The time for the ball to reach the top of its trajectory is the initial upward velocity divided by the acceleration due to gravity:
t = 19.6 m/s / 9.8 m/s² = 2 seconds. From the top of the trajectory, the ball will fall back down for (6 - 2) = 4 seconds.

We use the equation d = vi
t + 0.5 g
t2 to find the distance it falls back down: d = 0
t + 0.5
9.8 m/s²
42 = 78.4 meters. Hence, the height of the tower is 78.4 meters, option d.

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Rewritten by : Jeany

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