Asked by Beth
A ball is thrown directly downward, with an initial speed of 7.40 m/s, from a height of 31.0 m. After what time interval does the ball strike the ground?
I thought it would be 31 m / 7.40 = 4.18 seconds, but it's the wrong answer.
I thought it would be 31 m / 7.40 = 4.18 seconds, but it's the wrong answer.
Answers
Answered by
drwls
The acceleration of gravity, g = 9.81 m/s^2, has to have an effect on the answer.
Solve 0 = 31 - 7.40 t - (4.905)t^2
for t. Take the positive root of the quadratic equation.
t = [7.40 - sqrt (76.56 +588.6]/-9.81
= 1.87 s
Solve 0 = 31 - 7.40 t - (4.905)t^2
for t. Take the positive root of the quadratic equation.
t = [7.40 - sqrt (76.56 +588.6]/-9.81
= 1.87 s
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