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?

Question

I thought it would be 31 m / 7.40 = 4.18 seconds, but it's the wrong answer.

Answer 1

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