A football is thrown with an initial upward velocity component of 15.0m/s and a horizontal velocity component of 17.0m/s .

a:How much time is required for the football to reach the highest point in its trajectory?
b:How high does it get above the ground?
c:How much time after it is thrown does it take to return to its original height?
d:How far has the football traveled horizontally from its original position?

Xo = 17 m/s.

Yo = 15 m/s.

a. Tr = -Yo/g = -15/-9.8 = 1.53 s. = Rise Time.

b. h = 0.5g*Tr^2 = 4.9*1.53*2 = 11.47 m.

c. T = Tr+Tf = 1.53 + 1.53 = 3.06 s.

d. Dx = Xo*T = 17m/s. * 3.06 = 52.02 m.

Thanks

To solve these questions, we first need to find the vertical and horizontal components of the football's motion separately.

Given:
Initial upward velocity component (Vy) = 15.0 m/s
Horizontal velocity component (Vx) = 17.0 m/s

a: To find the time required for the football to reach the highest point in its trajectory, we need to consider the vertical motion. At the highest point, the vertical velocity component will be zero.

Using the equation:
Vy = Vy₀ + gt

Where:
Vy₀ = initial upward velocity component = 15.0 m/s
g = acceleration due to gravity = -9.8 m/s² (negative because it's in the opposite direction)
t = time

Rearranging the equation to solve for time (t):
t = (Vy - Vy₀) / g

Substituting the values:
t = (0 - 15.0) / -9.8
t = 1.53 seconds

Therefore, it takes 1.53 seconds for the football to reach the highest point in its trajectory.

b: To find how high the football gets above the ground, we need to calculate its maximum height, which occurs at the highest point in its trajectory.

Using the equation:
Vy² = Vy₀² + 2gΔy

Where:
Δy = vertical displacement or height
Vy = vertical velocity component (which is zero at the highest point)
Vy₀ = initial upward velocity component = 15.0 m/s
g = acceleration due to gravity = -9.8 m/s² (negative because it's in the opposite direction)

Rearranging the equation to solve for height (Δy):
Δy = (Vy² - Vy₀²) / (2g)

Substituting the values:
Δy = (0² - 15.0²) / (2 * -9.8)
Δy ≈ 11.53 meters

Therefore, the football reaches a peak height of approximately 11.53 meters above the ground.

c: To find the time it takes for the football to return to its original height, we simply need to double the time calculated in part b.

t total = 2 * t = 2 * 1.53 = 3.06 seconds

Therefore, it takes 3.06 seconds for the football to return to its original height.

d: To find how far the football has traveled horizontally from its original position, we need to consider the horizontal motion.

Using the equation:
distance = velocity × time

Where:
velocity = horizontal velocity component (Vx) = 17.0 m/s
time = total time of flight (which we already calculated as 3.06 seconds)

Substituting the values:
distance = 17.0 m/s × 3.06 s
distance ≈ 52.02 meters

Therefore, the football has traveled approximately 52.02 meters horizontally from its original position.

To answer these questions, we can use the equations of motion for projectile motion. These equations are based on the assumption that air resistance is negligible.

a) To find the time required for the football to reach the highest point, we need to consider the vertical motion. The initial upward velocity component is 15.0 m/s, and the acceleration due to gravity is -9.8 m/s² (negative because it acts downward). We can use the equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. At the highest point, the final velocity will be zero, so we have:

0 = 15.0 - 9.8t

Solving for t:

9.8t = 15.0
t = 15.0 / 9.8 ≈ 1.53 seconds

Therefore, it takes approximately 1.53 seconds for the football to reach the highest point in its trajectory.

b) To find the maximum height above the ground, we can use the equation:

hf = hi + vi*t + (1/2)*a*t²

where hf is the final height, hi is the initial height (which we assume to be 0), vi is the initial velocity, a is the acceleration, and t is the time. At the highest point, the final height will be the maximum height. Plugging in the values:

hf = 0 + 15.0*1.53 + (1/2)*(-9.8)*(1.53)²

hf ≈ 11.28 meters above the ground

Therefore, the football reaches a height of approximately 11.28 meters above the ground.

c) To find the time it takes for the football to return to its original height, we need to consider the entire vertical motion. We know that it took approximately 1.53 seconds to reach the highest point, so it should take another 1.53 seconds to come back down. Therefore, the total time is approximately 1.53 + 1.53 = 3.06 seconds.

So, it takes approximately 3.06 seconds for the football to return to its original height.

d) To find how far the football has traveled horizontally from its original position, we can use the equation:

dx = vix*t

where dx is the horizontal distance, vix is the initial horizontal velocity component, and t is the total time of flight. Plugging in the values:

dx = 17.0 * 3.06

dx ≈ 52.02 meters

Therefore, the football has traveled approximately 52.02 meters horizontally from its original position.