Anyone willing to give math help?
2015-09-14 2:55 pm
suppose a rock is dropped into a well 49 meters deep. when would it hit the ground? what is the velocity as it hits the ground? if the bottom of the well is lined with thick sand so that the rock does not bounce, what is the velocity of the rock just after it hits the bottom of the well?
回答 (1)
2015-09-14 3:09 pm
Take g = 9.8 m/s²
Take all downward quantities as positive.
Initial velocity, vₒ = 0 m/s
Acceleration, a = 9.8 m/s²
Displacement, s = 49 m
Time taken, t = ? s
Final velocity (velocity as it hits the ground), v = ? m/s
s = vₒt + (1/2)at²
49 = 0 + (1/2)(9.8)t²
t² = 10
Time taken, t = 3.2 s ...... (Ans)
v² = vₒ² + 2as
v² = 0 + 2(9.8)(49)
Velocity as it hits the ground, v = 31 m/s ...... (Ans)
The rock does not bounce.
Velocity of the rock just after it hits the bottom of the well = 0 m/s ...... (Ans)
Take all downward quantities as positive.
Initial velocity, vₒ = 0 m/s
Acceleration, a = 9.8 m/s²
Displacement, s = 49 m
Time taken, t = ? s
Final velocity (velocity as it hits the ground), v = ? m/s
s = vₒt + (1/2)at²
49 = 0 + (1/2)(9.8)t²
t² = 10
Time taken, t = 3.2 s ...... (Ans)
v² = vₒ² + 2as
v² = 0 + 2(9.8)(49)
Velocity as it hits the ground, v = 31 m/s ...... (Ans)
The rock does not bounce.
Velocity of the rock just after it hits the bottom of the well = 0 m/s ...... (Ans)
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