Final answer:
Mimi's softball went up 15.00625 meters high and left her bat at a speed of 17.15 m/s. Using the motion equations under gravity, we first find the time spent ascending (half of the total time) and then calculate the initial velocity. These values are then applied to find the maximum height.
Step-by-step explanation:
Solving for the Height and Initial Velocity of a Softball
To determine how high Mimi's softball went and how fast it left the bat, we can use the equations of motion for an object under the influence of gravity. Since the ball goes up and comes down in 3.5 seconds, the total time for the upward motion is half of that, which is 1.75 seconds. The acceleration due to gravity, which we'll denote as g, is approximately 9.8 m/s2 downwards.
The formula to find the maximum height (h) reached by the ball when thrown vertically is:
h = Vi * t - (1/2) * g * t2
Here, Vi is the initial velocity and t is the time taken to reach the maximum height. To find the initial velocity (Vi), we know that at the maximum height, the final velocity is 0 m/s. Hence, we use the formula:
Vf = Vi - g * t
As Vf = 0 at the highest point, we get:
Vi = g * t
After calculating Vi, we can substitute it back into the first formula to find h.
Let's solve part (A):
h = Vi * t - (1/2) * g * t2
Vi = 9.8 m/s2 * 1.75 s = 17.15 m/s
h = 17.15 m/s * 1.75 s - (1/2) * 9.8 m/s2 * (1.75 s)2
h = 30.0125 m - (1/2) * 9.8 m/s2 * 3.0625 s2
h = 30.0125 m - 15.00625 m = 15.00625 m
The ball went up 15.00625 meters high.
Now for part (B):
The initial velocity Vi is:
Vi = 9.8 m/s2 * 1.75 s = 17.15 m/s
Mimi's softball left her bat at a speed of 17.15 m/s.