A ball within an initial velocity of 5.55 m/s is shot at an angle of 30 degree above the horizontal using a launcher. The launcher is 1.11 high.
a. How long is the ball in the air, before it hits the ground?
b. How far from the launcher does the ball land on the ground?
a. using the free-fall equation ... 0 = -1/2 g t^2 + t sin(30º) + 1.11
... solve for t
b. distance = 5.5 cos(30º) * t ... use t from a.
To find the answers to these questions, we can use the equations of motion for projectile motion. Let's break down the problem step by step:
First, we need to find the vertical component of the initial velocity (Viy) and the horizontal component of the initial velocity (Vix).
Given:
- Initial velocity (V) = 5.55 m/s
- Launch angle (θ) = 30 degrees
- Launcher height (h) = 1.11 m
a. How long is the ball in the air, before it hits the ground?
Step 1: Find the vertical component of the initial velocity (Viy).
Using the formula:
Viy = V * sin(θ)
Viy = 5.55 m/s * sin(30°)
Viy = 2.775 m/s (rounded to three decimal places)
Step 2: Use the equation of motion for vertical motion to find the time of flight (t).
Using the formula:
t = (2 * Viy) / g
Where g is the acceleration due to gravity (approximately 9.8 m/s²).
t = (2 * 2.775 m/s) / 9.8 m/s²
t ≈ 0.564 s (rounded to three decimal places)
So, the ball is in the air for approximately 0.564 seconds.
b. How far from the launcher does the ball land on the ground?
Step 1: Find the horizontal component of the initial velocity (Vix).
Using the formula:
Vix = V * cos(θ)
Vix = 5.55 m/s * cos(30°)
Vix ≈ 4.808 m/s (rounded to three decimal places)
Step 2: Calculate the horizontal distance traveled (d) during the time of flight (t).
Using the formula:
d = Vix * t
d ≈ 4.808 m/s * 0.564 s
d ≈ 2.712 m (rounded to three decimal places)
So, the ball lands approximately 2.712 meters away from the launcher on the ground.