Explanation:To determine the initial velocity required for the second stone to reach the ground at the same instant as the first stone, we can analyze the motion of both stones separately.
For the first stone:
- Initial position (s1): 17.6 m (height of the bridge)
- Final position (sf1): 0 m (ground level)
- Distance traveled (d1): 2.70 m (distance the stone has fallen)
- Acceleration (a1): Acceleration due to gravity (-9.8 m/s^2, assuming downward as negative)
We can use the kinematic equation to find the time taken for the first stone to fall 2.70 m:
sf1 = s1 + v1t1 + (1/2) a1t1^2
0 = 17.6 + (1/2) (-9.8) t1^2
-17.6 = -4.9 t1^2
t1^2 = 3.5918
t1 ≈ 1.895 s
Now, for the second stone:
- Initial position (s2): 17.6 m (height of the bridge)
- Final position (sf2): 0 m (ground level)
- Distance traveled (d2): 17.6 m (height of the bridge)
- Acceleration (a2): Acceleration due to gravity (-9.8 m/s^2, assuming downward as negative)
- Initial velocity (v2): Unknown
Using the kinematic equation for motion with constant acceleration:
sf2 = s2 + v2t2 + (1/2) a2t2^2
0 = 17.6 + v2t2 + (1/2) (-9.8) t2^2
0 = 17.6 + v2t2 - 4.9 t2^2
Since the stones reach the ground at the same instant, the time taken by the second stone (t2) should be equal to the time taken by the first stone (t1):
t2 = t1 = 1.895 s
Substituting this value into the equation:
0 = 17.6 + v2(1.895) - 4.9(1.895)^2
0 = 17.6 + 1.895v2 - 17.8994025
Simplifying the equation:
1.895v2 = 17.8994025 - 17.6
1.895v2 = 0.2994025
v2 = 0.1578 m/s (approximately)
Therefore, to reach the ground at the same instant, the second stone should be given an initial velocity of approximately 0.1578 m/s.