A ball is thrown straight up and reaches a
maximum height of 6.71 m.
What was its initial speed? The accelera-
tion of gravity is 9.8 m/s2 .
Answer in units of m/s
please give right answer and explain how you got it, thanks !
To find the initial speed of a ball thrown straight up, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (which is 0 m/s at maximum height as the ball momentarily stops before falling back down)
vi = initial velocity (what we need to find)
a = acceleration due to gravity (which is -9.8 m/s^2 since it acts in the opposite direction to the ball's motion)
d = displacement (which is the maximum height, 6.71 m)
Substituting the given values into the equation:
0 = vi^2 + 2(-9.8)(6.71)
Simplifying:
0 = vi^2 - 2(9.8)(6.71)
Rearranging the equation:
vi^2 = 2(9.8)(6.71)
vi^2 = 2(65.54)
vi^2 = 131.08
Now, taking the square root of both sides to solve for vi:
vi = √131.08
vi ≈ 11.46 m/s
Therefore, the initial speed of the ball was approximately 11.46 m/s.