A quarterback throws a pass at an angle of 40 degrees above the horizontal with an initial speed of 20 m/s. The ball is caught by the receiver 2.55 seconds later. Determine horizontal velocity, vertical initial velocity, the distance ball was thrown, and maximum height of the ball. Can you do it? Show me...
Vo = 20m/s[40o].
Xo = 20*Cos40 = 15.3 m/s.
Yo = 20*sin40 = 12.9 m/s.
Dx = Xo*t = 15.3m/s * 2.55s. = 39 m.
h = Yo*Tr + 0.5g*Tr^2
Tr = 2.55/2 = 1.275 s. = Rise time.
g = -9.8 m/s^2.
Yes, I can help you with that. Let's break it down step by step.
Step 1: Determine the horizontal velocity (Vx):
The horizontal velocity remains constant throughout the motion. We can calculate it using the formula:
Vx = initial speed * cos(angle)
Given: initial speed = 20 m/s
angle = 40 degrees
Vx = 20 m/s * cos(40 degrees)
Vx ≈ 15.29 m/s (rounded to two decimal places)
Therefore, the horizontal velocity is approximately 15.29 m/s.
Step 2: Determine the vertical initial velocity (Vy):
The vertical initial velocity can be calculated using the formula:
Vy = initial speed * sin(angle)
Given: initial speed = 20 m/s
angle = 40 degrees
Vy = 20 m/s * sin(40 degrees)
Vy ≈ 12.86 m/s (rounded to two decimal places)
Therefore, the vertical initial velocity is approximately 12.86 m/s.
Step 3: Determine the distance the ball was thrown (horizontal range):
To calculate the distance, we can use the formula:
Range = horizontal velocity * time
Given: horizontal velocity = 15.29 m/s
time = 2.55 seconds
Range = 15.29 m/s * 2.55 seconds
Range ≈ 39.05 meters (rounded to two decimal places)
Therefore, the ball was thrown approximately 39.05 meters.
Step 4: Determine the maximum height of the ball:
To calculate the maximum height, we can use the formula:
Max height = (vertical initial velocity)^2 / (2 * g)
Given: vertical initial velocity = 12.86 m/s
g = acceleration due to gravity = 9.8 m/s^2
Max height = (12.86 m/s)^2 / (2 * 9.8 m/s^2)
Max height ≈ 8.71 meters (rounded to two decimal places)
Therefore, the maximum height of the ball is approximately 8.71 meters.
To summarize:
- Horizontal velocity = approximately 15.29 m/s
- Vertical initial velocity = approximately 12.86 m/s
- Distance ball was thrown (horizontal range) = approximately 39.05 meters
- Maximum height of the ball = approximately 8.71 meters
Sure! I can help you with that. We can solve this problem by breaking it down into different parts and using some basic physics equations.
1. First, let's find the horizontal velocity. The horizontal velocity remains constant throughout the projectile motion because there are no horizontal forces acting on the ball. So, the initial horizontal velocity is the same as the final horizontal velocity.
Horizontal velocity (Vx) = Initial velocity (V) * Cosine of the launch angle (θ)
Vx = 20 m/s * cos(40°)
Vx ≈ 20 m/s * 0.766
Vx ≈ 15.32 m/s
Therefore, the horizontal velocity of the ball is approximately 15.32 m/s.
2. Next, let's find the vertical initial velocity. The vertical velocity changes due to the acceleration of gravity. The initial vertical velocity can be found using the sine of the launch angle.
Vertical initial velocity (Vy) = Initial velocity (V) * Sine of the launch angle (θ)
Vy = 20 m/s * sin(40°)
Vy ≈ 20 m/s * 0.643
Vy ≈ 12.86 m/s
Therefore, the vertical initial velocity of the ball is approximately 12.86 m/s.
3. To find the distance the ball was thrown, we can use the horizontal velocity and the time it takes for the ball to be caught.
Distance (d) = Horizontal velocity (Vx) * Time (t)
d = 15.32 m/s * 2.55 s
d ≈ 39.16 m
Therefore, the distance the ball was thrown is approximately 39.16 meters.
4. Finally, let's find the maximum height of the ball. At the maximum height, the vertical velocity becomes zero, and the ball starts descending.
Using the vertical motion equation:
Vy = Vertical initial velocity (Vy) - (Acceleration due to gravity * Time (t))
0 = 12.86 m/s - (9.8 m/s^2 * t)
9.8t = 12.86
t ≈ 1.31 s
Now, let's find the maximum height (h) using this time:
h = Vertical initial velocity (Vy) * Time (t) - (0.5 * Acceleration due to gravity * Time (t))^2
h = 12.86 m/s * 1.31 s - (0.5 * 9.8 m/s^2 * 1.31 s)^2
h ≈ 8.29 m
Therefore, the maximum height of the ball is approximately 8.29 meters.
To summarize:
- Horizontal velocity (Vx) ≈ 15.32 m/s
- Vertical initial velocity (Vy) ≈ 12.86 m/s
- Distance thrown (d) ≈ 39.16 meters
- Maximum height (h) ≈ 8.29 meters