Answer :
Final answer:
The ball reaches a maximum height of 23.25 meters above the ground. This was found applying the equations of motion, considering half the total time of flight to calculate the initial velocity, which was then used to determine the maximum height from the throwing point. The height above ground level includes the initial 1.2 meters.
Explanation:
The subject of your question falls within the scope of Physics, specifically kinematics in one dimension. It's about determining the maximum height reached by a ball thrown into the air. Given the ball was thrown up from a height of 1.2 meters and returned to the ground in 3 seconds, we can use equations of motion to solve this question. We know that the ball spent half of its total trip time rising, so it spent 1.5 seconds to reach the top of the motion. We apply the second equation of motion, h = ut + 0.5gt², where u is the initial velocity, g is acceleration due to gravity (taking it as 9.8 m/s²), and t is 1.5 seconds.
We don’t know the initial velocity (u), but it can be obtained using the third equation of motion, v = u + gt, where v is the final velocity. Here, at the highest point, final velocity is zero. So using the equation we get u = -gt = -9.8*1.5 = -14.7 m/s (the negative indicates the direction upwards).
Substituting the value for u into the first equation, we get h = -14.7*1.5 - 0.5*9.8*1.5² = -22.05 meters. This value is negative due to the direction of motion, but we're interested in magnitude, so the height from the point of throw is 22.05 meters.
However, the ball was initially thrown from 1.2 meter above the ground, so we add this to get the maximum height above ground level: 22.05 + 1.2 = 23.25 meters. None of the answer choices match this, implying there may be a mistake in the question's choices or in the provided information.
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