a football is thrown to a moving receiver .the football leaves the quarterback's hands 1.75m above the ground with a velocity of 17.0m/s[25deg]if the receiver starts 12.0m away from the quarterback along the line of flight of the ball when is is thrown , what constant velocity must she have to get to the ball at the instant it is 1.75 m above the ground ?
Vo = 17m/s[25o]
Xo = 17*cos25 = 15.4 m/s.
Yo = 17*sin25 = 7.2 m/s.
Y = Yo + g*t = 0 @ max. Ht.
Tr = -Yo/ g = -7.2/-9.8 = 0.73 s. = Rise
time.
Tf = Tr = 0.73 s. = Fall time.
D = Xo*(Tr+Tf) = 15.4 * 1.46 = 22.5 m.
V=(D-12)/(Tr+Tf)=(22.5-12)/1.46=7.19m/s
Well, that's quite a toss! Let's see if we can calculate the constant velocity the receiver needs to reach the ball at the right moment.
First, we need to break the initial velocity of the football into its vertical and horizontal components. The vertical component can be found using the angle of 25 degrees: 17.0 m/s * sin(25) = 7.1 m/s. The horizontal component is found using the cosine function: 17.0 m/s * cos(25) ≈ 15.3 m/s.
Now, let's focus on the horizontal motion. We know that the receiver starts 12.0 m away from the quarterback, so they need to cover the remaining distance of 12.0 m. The time it takes is given by the equation:
time = distance / velocity
time = 12.0 m / 15.3 m/s ≈ 0.78 s
So, the receiver needs to have a constant velocity of 12.0 m / 0.78 s ≈ 15.4 m/s to reach the ball at the instant it is 1.75 m above the ground. That's quite the speedy receiver! I hope they have their running shoes on.
To find the constant velocity the receiver must have to reach the ball at the instant it is 1.75m above the ground, we can break down the problem into horizontal and vertical components.
Let's first find the time it takes for the ball to reach a height of 1.75m.
We know the initial vertical velocity (Viy) is 17.0 m/s * sin(25°), and the acceleration in the vertical direction is -9.8 m/s^2 (due to gravity). We can use the kinematic equation:
Vf = Vi + at
where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time.
Since the ball's final velocity at the highest point is 0 m/s (ball momentarily comes to rest at the highest point), we can rearrange the equation as follows:
0 = Viy - gt
We can solve this equation for t:
t = Viy / g
Plugging in the values:
t = (17.0 m/s * sin(25°)) / 9.8 m/s^2
t ≈ 1.77 s
Now that we have the time it takes for the ball to reach a height of 1.75m, we can find the horizontal distance the receiver needs to cover in that time.
Using the equation:
distance = velocity * time,
we can find the horizontal distance traveled by the ball:
distance = 17.0 m/s * cos(25°) * t
distance = 17.0 m/s * cos(25°) * 1.77 s
distance ≈ 28.9 m
So, the receiver must reach a distance of approximately 28.9m from the starting point of the ball to catch it at the instant it is 1.75m above the ground. The constant velocity required for the receiver to cover this distance in 1.77 seconds is approximately 28.9m / 1.77s ≈ 16.3 m/s.
To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion.
First, let's analyze the vertical motion. We know that the football leaves the quarterback's hands 1.75m above the ground and the receiver wants to catch it at the same height. The vertical motion can be described using the equation:
𝑑 = 𝑣₀𝑡 + 0.5𝑎𝑡²,
where 𝑑 is the vertical displacement, 𝑣₀ is the initial vertical velocity, 𝑡 is the time, and 𝑎 is the acceleration due to gravity (-9.8 m/s²).
In this case, 𝑑 = 1.75 m, 𝑣₀ = 0 m/s (since the receiver wants to catch it at the same height), and 𝑎 = -9.8 m/s². Substituting these values, we can solve for 𝑡.
1.75 = 0(𝑡) + 0.5(-9.8)(𝑡)²
1.75 = -4.9𝑡²
𝑡² = 1.75 / -4.9
𝑡² ≈ -0.357
Since time cannot be negative, we discard the negative value. Therefore, 𝑡 is imaginary, which means it is not possible for the receiver to catch the ball at the same height as it leaves the quarterback's hands.
Now let's move on to the horizontal motion. We know that the receiver starts 12.0m away from the quarterback along the line of flight of the ball. The horizontal motion can be described using the equation:
𝑑 = 𝑣₀𝑡,
where 𝑑 is the horizontal displacement, 𝑣₀ is the initial horizontal velocity, and 𝑡 is the time.
In this case, 𝑑 = 12.0 m, and 𝑡 is the same value we obtained earlier (even though it was imaginary). We can solve for 𝑣₀.
12.0 = 𝑣₀(𝑡)
𝑣₀ = 12.0 / 𝑡
Since 𝑡 is imaginary, there is no valid solution for 𝑣₀. Therefore, it is not possible for the receiver to have a constant velocity to catch the ball at the instant it is 1.75 m above the ground, given the initial conditions and constraints provided in the question.