A baseball is thrown, leaving the baseball player’s hand 6.0 ft about the ground and at a 45 degree angle with respect to the horizon. The initial speed is 80 ft/s. Find the maximum height attained, the length of time the ball stays in the air, and the horizontal distance where the ball hits the ground. (assume air resistance is negligible).

Question

A baseball is thrown, leaving the baseball player’s hand 6.0 ft about the ground and at a 45 degree angle with respect to the horizon. The initial speed is 80 ft/s. Find the maximum height attained, the length of time the ball stays in the air, and the horizontal distance where the ball hits the ground. (assume air resistance is negligible).

Answer 1

YOu know the initial vertical component, and the initial horizontal component.

final height=initialheight+vovertical*time-1/2 g time^2
solve for time...this is time in the air.
Now at max height, vvertical will be zero
0=vovertical-g*time solve for time max height, then use that time in the h equation to determine height.

Answer 2

To find the maximum height attained, we need to consider the vertical motion of the ball. We can start by breaking down the initial velocity of the ball into its vertical and horizontal components.

Given that the initial velocity is 80 ft/s and the angle with respect to the horizon is 45 degrees, we can calculate the initial vertical velocity using trigonometry.

Vertical component of initial velocity (V_y) = initial velocity (V) * sine(angle)

V_y = 80 ft/s * sine(45 degrees)
V_y = 80 ft/s * 0.7071
V_y = 56.568 ft/s

We can now use the formula for vertical motion to find the maximum height attained by the ball:

Max height (H) = (V_y^2) / (2 * g)

Where g is the acceleration due to gravity, approximately 32.2 ft/s^2.

H = (56.568 ft/s)^2 / (2 * 32.2 ft/s^2)
H ≈ 179.473 ft

So, the maximum height attained by the ball is approximately 179.473 ft.

To find the length of time the ball stays in the air, we can use the formula for the total time of flight:

Time of flight (T) = (2 * V_y) / g

T = (2 * 56.568 ft/s) / 32.2 ft/s^2
T ≈ 3.507 s

So, the ball stays in the air for approximately 3.507 seconds.

To find the horizontal distance where the ball hits the ground, we can use the formula for horizontal motion:

Horizontal distance (D) = initial velocity (V) * cosine(angle) * time of flight (T)

D = 80 ft/s * cosine(45 degrees) * 3.507 s
D = 80 ft/s * 0.7071 * 3.507 s
D ≈ 198.395 ft

So, the ball hits the ground at a horizontal distance of approximately 198.395 ft.