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Answer :
Final answer:
The initial velocity of a ball thrown vertically upwards and returning after 6 seconds, with gravity at 9.8 m/s², is 29.4 m/s, corresponding with option B.
Explanation:
The student's question involves calculating the initial velocity of a ball thrown vertically upwards, given that it returns to the thrower's hand after 6 seconds. Assuming the acceleration due to gravity (g) as 9.8 m/s², the required initial velocity can be found using the kinematic equation for uniformly accelerated motion.
For the ball to return in 6 seconds, it takes half of that time, namely 3 seconds, to reach the apex of its trajectory where its velocity becomes 0. Therefore, we can use the velocity-time relation for uniformly accelerated motion, which is:
v = u + at
At the apex, v = 0 (the final velocity), a = -9.8 m/s² (acceleration due to gravity acting downwards), and t = 3 s (the time taken to reach the apex).
Plugging the values into the equation yields:
0 = u - (9.8 m/s²)(3 s)
u = (9.8 m/s²)(3 s)
u = 29.4 m/s²
Thus, the initial velocity with which the ball was thrown up is 29.4 m/s, which corresponds to option B.
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