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.
