A third baseman makes a throw to first base 39.8 m away. The ball leaves his hand with a speed of 43.0 m/s at a height of 1.4 m from the ground and making an angle of 23.7 o with the horizontal. How high will the ball be when it gets to first base?

To find the height of the ball when it gets to first base, we need to first determine the time it takes for the ball to reach there.

To find the time, we can use the horizontal component of the ball's velocity. The horizontal component can be found using the formula:

Vx = V * cos(theta)

Where:
Vx = horizontal component of velocity
V = initial velocity (43.0 m/s in this case)
theta = angle with the horizontal (23.7 degrees in this case)

Using this formula, we can find Vx:

Vx = 43.0 m/s * cos(23.7 degrees)
Vx ≈ 39.12 m/s

Next, we can calculate the time of flight using the formula:

t = distance / Vx

Where:
t = time
distance = distance to first base (39.8 m in this case)
Vx = horizontal component of velocity (39.12 m/s in this case)

Using this formula, we can find t:

t = 39.8 m / 39.12 m/s
t ≈ 1.017 seconds

Now that we know the time of flight, we can find the vertical distance traveled by the ball using the formula:

y = V * sin(theta) * t - (1/2) * g * t^2

Where:
y = vertical distance
V = initial velocity (43.0 m/s in this case)
theta = angle with the horizontal (23.7 degrees in this case)
t = time of flight (1.017 seconds in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Using this formula, we can find y:

y = 43.0 m/s * sin(23.7 degrees) * 1.017 s - (1/2) * 9.8 m/s^2 * (1.017 s)^2
y ≈ 4.94 m

Therefore, the ball will be approximately 4.94 meters above the ground when it gets to first base.