Answer: after 1.75 seconds
Explanation:
The only force acting on the ball is the gravitational force, so the acceleration will be:
a = -9.8 m/s^2
the velocity can be obtained by integrating over time:
v = -9.8m/s^2*t + v0
where v0 is the initial velocity; v0 = -7.95 m/s.
v = -9.8m/s^2*t - 7.95 m/s.
For the position we integrate again:
p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0
where p0 is the initial position: p0 = 29m
p = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Now we want to find the time such that the position is equal to zero:
0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Then we solve the Bhaskara's equation:
[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]
Then the solutions are:
t = (7.95 + 25.1)/(-9.8) = -3.37s
t = (7.95 - 25.1)/(-9.8) = 1.75s
We need the positive time, then the correct answer is 1.75s
