a footbal is thrown at an angle of 35 reltive to the ground at a speed of 25m/s. the football is then caught 38m from the throwner. you can assume the height that the ball thrown at is the same as the height it was caught. how long was the football in the air ?
Vo = 25m/s @ 35o
Xo = 25*cos35 = 20.48 m/s.
Dx = Xo*t = 38 m.
t = 38/Xo = 38/20.48 = 1.86 s.
To determine the time the football was in the air, we need to find the time it takes for the football to travel the horizontal distance of 38m.
First, let's resolve the initial velocity of the football into its horizontal and vertical components:
Horizontal component: Vx = V * cos(θ)
Vertical component: Vy = V * sin(θ)
Given:
- Angle of 35 degrees: θ = 35 degrees
- Speed of 25 m/s: V = 25 m/s
Using these values, we can calculate the horizontal and vertical components of velocity:
Vx = 25 m/s * cos(35 degrees)
Vy = 25 m/s * sin(35 degrees)
Next, we need to find the time it takes for the football to reach a horizontal distance of 38m. The horizontal distance traveled by an object can be calculated using the formula:
Distance = Velocity * Time
Since the football's initial horizontal velocity is constant (there is no horizontal acceleration), we can solve for time using the formula:
Time = Distance / Velocity
Plugging in the given values:
Time = 38 m / Vx
Now we can calculate the time the football was in the air:
Time = 38 m / (25 m/s * cos(35 degrees))
Calculating the value using a calculator:
Time ≈ 38 m / (25 m/s * 0.819)
So, the football was in the air for approximately:
Time ≈ 1.84 seconds