Two Dimensional Projectile Question:

Determine the height reached by a baseball if it is released with a velocity of 17.0 m/s [at an angle of 20 degrees]

It rises for a time T such that

Vo sin 20 = g T
T = Vo sin 20/g = 0.5933 s

The height reached is
(Vo sin20)*T/2 = (Vo sin20)^2/(2 g)

To determine the height reached by the baseball, we can break the problem into two separate components: horizontal and vertical motion. Let's first calculate the time it takes for the baseball to reach the highest point of its trajectory.

Step 1: Calculate the vertical initial velocity (Vy) and horizontal initial velocity (Vx):
Given:
- Initial velocity (V) = 17.0 m/s
- Launch angle (θ) = 20 degrees

To calculate the vertical initial velocity (Vy):
Vy = V * sin(θ)
Vy = 17.0 m/s * sin(20°)
Vy ≈ 5.8 m/s

To calculate the horizontal initial velocity (Vx):
Vx = V * cos(θ)
Vx = 17.0 m/s * cos(20°)
Vx ≈ 16.1 m/s

Step 2: Calculate the time to reach maximum height:
The time it takes for an object to reach maximum height is half of the total time it takes to complete its projectile motion.

t_max = Vy / g
Where g is the acceleration due to gravity (approximately 9.8 m/s²).

t_max = 5.8 m/s / 9.8 m/s²
t_max ≈ 0.59 seconds

Step 3: Calculate the maximum height (H):
The maximum height is reached when the vertical component of velocity becomes zero.

H = Vy² / (2 * g)
H = (5.8 m/s)² / (2 * 9.8 m/s²)
H ≈ 1.7 meters

Therefore, the height reached by the baseball is approximately 1.7 meters.