A soccer ball is kicked with a speed of 9.95m/s at an angle of 35.0∘ above the horizontal.If the ball lands at the same level from which it was kicked, how long was it in the air?
velocity up ;9.95sin35
hf=hi+vi*t-1/2 g t^2
hf, hi=0
solve for time t.
To find the time the soccer ball was in the air, we can start by breaking down its initial velocity into its horizontal and vertical components.
Given:
Initial speed (v) = 9.95 m/s
Launch angle (θ) = 35.0 degrees
The vertical component of the initial velocity can be calculated as:
Vy = v * sin(θ)
= 9.95 m/s * sin(35.0 degrees)
≈ 5.69 m/s
The time of flight (t) can be found using the vertical component of the velocity. At its highest point, the vertical component of velocity becomes zero, and the only force acting on the ball is gravity. We can use the equation:
Vy = -gt
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation, we get:
t = -Vy / g
Substituting the given values, we have:
t = -5.69 m/s / (9.8 m/s^2)
≈ -0.581 seconds
Since time cannot be negative, we can take the absolute value of the result:
t ≈ 0.581 seconds
Therefore, the soccer ball was in the air for approximately 0.581 seconds.
To determine how long the soccer ball was in the air, we need to analyze the projectile motion of the ball. This involves breaking down the motion into its horizontal and vertical components.
Given information:
- Initial velocity (speed): 9.95 m/s
- Launch angle: 35.0° above the horizontal
Step 1: Analyzing the vertical motion
Since the ball lands at the same level it was kicked, we know that the vertical displacement is zero. Therefore, the time of flight can be determined using the vertical motion.
The vertical motion of the ball can be analyzed using the equation:
Δy = V₀y * t + (1/2) * g * t²
Where:
- Δy is the vertical displacement (which is zero in this case)
- V₀y is the initial vertical velocity
- g is the acceleration due to gravity (approximately 9.8 m/s²)
- t is the time of flight
Since the ball is kicked at an angle above the horizontal, we first need to determine the initial vertical velocity:
V₀y = V₀ * sin(θ)
V₀ is the initial velocity (9.95 m/s) and θ is the launch angle (35.0°). Therefore:
V₀y = 9.95 m/s * sin(35.0°)
Step 2: Solving for time of flight
Since Δy = 0, the equation becomes:
0 = V₀y * t + (1/2) * g * t²
Now we can substitute the values and solve for t:
0 = (9.95 m/s * sin(35.0°)) * t + (1/2) * (9.8 m/s²) * t²
This is a quadratic equation in the form of at² + bt + c = 0, where a = (1/2) * g, b = V₀y, and c = 0. We can solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = (1/2) * (9.8 m/s²), b = (9.95 m/s * sin(35.0°)), and c = 0.
Solving the quadratic equation will give us two solutions, but we only need the positive solution since time cannot be negative:
t = (-b + √(b² - 4ac)) / (2a)
Step 3: Calculate the time of flight
Now, we can substitute the values and calculate the time of flight:
t = [-(9.95 m/s * sin(35.0°)) + √((9.95 m/s * sin(35.0°))² - 4 * (1/2) * (9.8 m/s²) * 0)] / [2 * (1/2) * (9.8 m/s²)]
Calculate the values inside the square root and simplify the equation to find the time of flight.
Please note that due to rounding or variations in the initial parameters, your final answer may slightly differ.
0 = 0 + Vi t - 4.9 t^2
Vi = 9.95 sin 35 = 5.71
so
0 = 5.71 t - 4.9 t^2
t = 0 or t = 1.16
t = 0 was zero height when it started
t = 1.16 seconds is zero height when it landed