a football is kicked with a speed of 15m/s
at an angle of 35 degree from the ground.
a. how high is the ball from the ground at t=0.400s?
b. how fast is it moving(s) after being kicked?
c. at what t-values is the ball 2m from the ground?
Vo = 15m/s @ 35deg.
Xo = hor. = 15cos35 = 12.29m/s.
Yo = ver. = 15sin35 = 8.60m/s.
a. h = Yo*t + 0.5g*t^2,
h = 8.60*0.4 - 4.9*(0.4)^3 = 2.66m.
b. Vo = 15m/s.
c. d = Yo*t - 4.9*t^2 = 2m,
8.60t - 4.9t^2 = 2,
-4.9t^2 + 8.6t - 2 = 0.
Tr = o.27594s. = Time to rise to 2m above ground. Use Quadratic Formula and choose shortest time.
hmax = (Yf^2 - Yo^2) / 2g,
hmax=(0 - (8.60)^2) / -19.6 = 3.77m.
d = Yo*t + 4.9t^2 = 3.77 - 2 = 1.77m.
0 + 4.9t^2 = 1.77,
t^2 = 0.361,
Tf = 0.60s = Time to fall to 2m. above ground.
a. How high is the ball from the ground at t=0.400s?
Well, let's calculate it using a bit of physics and a whole lot of imagination. You see, when the football is kicked, it goes on a mighty adventure through the air. And after a certain time of 0.400 seconds, it reaches a certain height.
To determine that height, we can use the formula for vertical displacement:
𝑦 = 𝑣𝑜𝑦𝑡 − (1/2)𝑔𝑡²
Where:
𝑦 = vertical displacement
𝑣𝑜𝑦 = initial vertical velocity
𝑔 = acceleration due to gravity (approximately 9.8m/s²)
𝑡 = time
Now, since the football was kicked at an angle of 35 degrees, we can split its initial velocity into vertical and horizontal components. The vertical component can be calculated using 𝑣𝑜𝑦 = 𝑣𝑜 × sin(𝜃), where 𝑣𝑜 is the initial speed (15m/s) and 𝜃 is the angle (35 degrees).
So, 𝑣𝑜𝑦 = 15m/s × sin(35°).
Plug in all the values, and you'll get the height of the ball from the ground at t = 0.400 seconds. But beware – this height may not be as high as a trapeze artist, but it should still be pretty decent. Enjoy the calculations!
b. How fast is it moving after being kicked?
Well, since the football was kicked with a speed of 15m/s, I can confidently say that it's moving pretty darn fast! It's zooming through the air like a cheetah on roller skates. But if you want a more accurate answer, you'll need to specify how long after being kicked you're asking about. I can't predict the future, you know!
c. At what t-values is the ball 2m from the ground?
Ah, the elusive 2m mark! You want to know when the ball is at that height, huh? Well, fear not, for I have the solution! By using the same vertical displacement formula as before (𝑦 = 𝑣𝑜𝑦𝑡 − (1/2)𝑔𝑡²), we can set 𝑦 equal to 2m and solve for 𝑡.
But hang on just a second! Remember that the vertical component of initial velocity was calculated as 𝑣𝑜𝑦 = 15m/s × sin(35°). So, plug that value into the equation, set 𝑦 equal to 2m, and you'll find the t-values when the ball is 2m from the ground. It might just be hanging out there, contemplating the deeper meaning of life.
To answer these questions, we first need to break down the initial velocity of the football into its horizontal and vertical components.
Given:
Initial speed (v₀) = 15 m/s
Angle (θ) = 35 degrees
a. To find the height of the ball at t = 0.400s, we can use the formula for vertical displacement:
h = v₀ * sin(θ) * t - (1/2) * g * t²
where:
v₀ is the initial speed,
θ is the angle of the trajectory,
t is the time, and
g is the acceleration due to gravity.
Since we are given v₀ and θ, we can substitute those values into the formula:
h = 15 * sin(35) * 0.400 - (1/2) * 9.8 * 0.400²
Calculating the above expression gives us the height of the ball at t = 0.400s from the ground.
b. To find the speed of the ball after being kicked, we can use the formula for the horizontal component of velocity:
vx = v₀ * cos(θ)
where vx is the horizontal component of velocity.
Substituting the given values, we can calculate vx.
c. To find the t-values when the ball is 2m from the ground, we can use the equation for vertical displacement:
h = v₀ * sin(θ) * t - (1/2) * g * t²
Rearranging the equation, we get:
t = (v₀ * sin(θ) ± √((v₀ * sin(θ))² + 2 * (1/2) * g * 2)) / g
Now, substitute the given values (v₀, θ, g, and h) into the equation to find the corresponding t-values. Remember that there may be two solutions, since the ball can be at the given height at two different points in its trajectory.
To answer these questions, we can use the principles of projectile motion. Projectile motion refers to the motion of objects that are launched into the air and move under the influence of gravity.
a. To determine the height of the ball from the ground at t = 0.400 s, we need to consider the vertical component of its motion. We can calculate the height using the equation:
h = h0 + (V0y * t) - (0.5 * g * t^2)
where:
- h is the height from the ground,
- h0 is the initial height (which is zero in this case),
- V0y is the initial vertical velocity of the ball (obtained from the initial speed and launch angle),
- t is the time, and
- g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
First, we need to calculate the initial vertical velocity of the ball (V0y). We can use the following equation:
V0y = V0 * sin(θ)
where:
- V0 is the initial speed of the ball (15 m/s in this case),
- θ is the launch angle (35 degrees in this case).
After calculating V0y, we can substitute it into the first equation and solve for h.
b. To determine how fast the ball is moving after being kicked, we need to consider the horizontal component of its motion. Since there is no horizontal acceleration (assuming no air resistance), the initial horizontal velocity (V0x) will remain constant. We can calculate it using the equation:
V0x = V0 * cos(θ)
where:
- V0 is the initial speed of the ball (15 m/s in this case),
- θ is the launch angle (35 degrees in this case).
The speed of the ball can be obtained by taking the square root of the sum of the squares of the horizontal velocity and vertical velocity at any given time.
c. To determine the time values at which the ball is 2m from the ground, we can again use the equation for the height of the ball. We set h = 2 and solve for t. However, in this case, we should consider both the upward and downward motion of the ball. So, we will have two different time values.
Now, let's calculate the answers to your questions:
a. To calculate the height of the ball from the ground at t = 0.400 s, plug the given values into the equation:
- h0 = 0 m (initial height)
- V0 = 15 m/s (initial speed)
- θ = 35 degrees (launch angle)
- t = 0.400 s
b. To calculate the speed at any given time, we need to calculate the horizontal (Vx) and vertical (Vy) components of the velocity separately. Then, we can find the magnitude of the velocity using the equation: v = √(Vx^2 + Vy^2). Plug the given values into the equations:
- V0 = 15 m/s (initial speed)
- θ = 35 degrees (launch angle)
c. To calculate the time values at which the ball is 2m from the ground, we need to solve the equation for t when h = 2. Plug the given values into the equation:
- h = 2 m
- V0 = 15 m/s (initial speed)
- θ = 35 degrees (launch angle)