Physics
Consider the position of a ball thrown down
with an initial speed of 17 m/s.What will be its position after 2.6 s? Let
the initial position be 0. The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of m
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Answers
The position of the ball after 2.6 seconds will be 81.88 m. This can be calculated using the equation distance = initial velocity x time + 1/2 x acceleration x time2. In this equation, the initial velocity is 17 m/s, the initial position is 0, the acceleration is 9.8 m/s2, and the time is 2.6 s. Therefore, the position of the ball after 2.6 seconds will be 17 x 2.6 + 0.5 x 9.8 x (2.6)2 = 81.88 m.
