**Speed of a Softball**

Part A = 18.77 m/s

Part B = 31.50 °

Part C = 16.0, -8.82 m/s

Part D = 30.4, 0.932 m

**Solution Below:**

_{0}and at an angle θ from the horizontal. Immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity V = 7.000 m/s, for a time t = 2.000 s. He then catches the ball at the same height at which it left the bat. The third baseman was initially L = 18.00 m from the location where the ball was hit at home plate.

Part A

_{0}. Use g = 9.807 m/s

^{2}for the magnitude of the acceleration due to gravity.

Express the initial speed in units of meters per second to four significant figures.

Since it takes 2 seconds for the ball to go up and then back down, we know that it takes half that time (1 second) for the ball to reach it’s maximum height. At the maximum height, the ball’s y-velocity is zero. The initial velocity in the y direction can be found with the following formula:

v_{y} = g / t

v_{y} = 9.807 / 1

v_{y} = 9.807 m/s

For the x velocity, the third baseman runs for 2 seconds at a speed of 7 m/s. This means he runs a total of 14 m. Add that to the starting distance from where the ball was hit from (18 m) for a total x-distance of 32 m. The ball covered this distance in 2 seconds, so it must have been going 16 m/s in the x direction. Now add the x and y components of the velocity using the Pythagorean theorem:

v = sqrt(x^2 + y^2)

v = sqrt(16^2 + 9.807^2)

v = sqrt(352.177)

v = 18.77 m/s

18.77 m/s

Part B

Express your answer in degrees to four significant figures.

With a velocity of 18.77, all we have to do is find the angle that gives the corresponding x or y velocity from Part A. Let’s just use the y velocity:

v * sin(θ) = v_{y}

18.77 * sin(θ) = 9.807

sin(θ) = 0.52248

θ = 31.50 °

31.50 °

Part C

Use the notation v

_{x}, v

_{y}, an ordered pair of values separated by commas. Express your answer in units of meters per second to three significant figures.

The only component of v that will change from start to finish is the y velocity. Since it takes 1 second for the ball to reach its maximum height (Part A), it also takes 1 second for it to fall back down. This means that 0.1 s before the ball is caught, it will have been falling for 0.9 s, and it’s speed in the y direction is (0.9 * 9.807) = -8.8263 m/s. Note that this should be a negative, since the ball is falling downwards. The x component will not change. We found in part A that it is 16 m/s

16.0, -8.82 m/s

Part D

Use the notation v

_{x}, v

_{y}, an ordered pair of values separated by commas. Express your answer in units of meters per second to three significant figures.

Since the ball will have been moving for 1.9 s, we know it will have traveled (16.0 * 1.9) = 30.4 m in the x-direction. How far will it be above the ground (in the y-direction)? We earlier calculated that the ball will be moving at 8.8263 m/s in the y direction 0.1 seconds before it’s caught (Part C). Once it is caught, it will be moving at 9.807 m/s, since it will have fallen all the way back down. So we can solve using the following formula:

Vf^2 = Vi^2 + 2a(Δy)

9.807^2 = 8.8263^2 + 2 * 9.807 * Δy

96.2 = 77.9 + 19.614 * Δy

Δy = 18.3 / 19.614

Δy = 0.933 m

30.4, 0.932 m