If a ball is hit with an initial velocity of 110 feet per second at an angle of 45 degrees from an initial height of 2 feet, how long will it be in the air?The following parametric equations are used to represent the location of the ball after t seconds.
So you do not give me the equations
and
you make me work in silly old units
oh well
u = x velocity = 110 cos 45 forever
Vi = initial up velocity = 110 sin 45
g = 32 ft/second^2 in old English units
v = Vi - g t = 110 sin 45 - 32 t
h = Hi + Vi t - (g/2)t^2 = 2 + 110 sin 45 * t - 16 t^2
when will h be zero ?
0 = 2 + 77.8 t - 16 t^2
solve quadratic
https://www.mathsisfun.com/quadratic-equation-solver.html
about 4.9 seconds
now if you want it in parametric form use x = u t
where u = 110 cos 45
so t = x /77.8
then
h = Hi + Vi t - (g/2)t^2 = 2 + 110 sin 45 * t - 16 t^2
or
h = 2 + 110 sin 45 * t - 16 t^2
or
h = 2 + 77.8 t -16 t^2
becomes
h = 2 + 77.8 (x/77.8) - 16 (x^2/77.8^2)
h = 2 + x - 16 x^2/77.8^2
that parabola gives you a quadratic for x when h = 0 then go back and get t
To find the time when the ball will be in the air, we can utilize the vertical component of the ball's motion. The equation for the vertical position of the ball can be represented as:
y(t) = h0 + v0yt - (1/2)gt^2
where:
t is the time in seconds
y(t) is the vertical position of the ball at time t
h0 is the initial height of the ball
v0y is the vertical component of the initial velocity
g is the acceleration due to gravity (~32.2 ft/s^2)
In this case, h0 = 2 ft and the initial velocity v0 = 110 ft/s at an angle of 45 degrees. Since the initial velocity is given at an angle, we need to decompose it into horizontal and vertical components.
v0y = v0 * sin(θ)
where θ is the launch angle.
v0y = 110 ft/s * sin(45°)
v0y = 110 ft/s * 0.7071
v0y ≈ 77.78 ft/s
We can now rewrite the vertical position equation with the given values:
y(t) = 2 + 77.78t - (1/2)(32.2)t^2
To determine how long the ball will be in the air, we need to find the time when the ball hits the ground, i.e., when y(t) = 0.
0 = 2 + 77.78t - (1/2)(32.2)t^2
This equation is a quadratic equation that can be solved for t. We can use the quadratic formula to find the solutions:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = -(1/2)g, b = v0y, and c = h0.
Substituting these values into the quadratic formula:
t = (-(77.78) ± √((77.78)^2 - 4(-16.1)(2))) / 2(-16.1)
Now, we can calculate the values inside the square root:
√((77.78)^2 - 4(-16.1)(2))
√(6050.08 + 129.92)
√6180
t ≈ (-(77.78) ± √6180) / (-32.2)
Calculating the values inside the square root gives two solutions:
t ≈ (-(77.78) + √6180) / (-32.2)
t ≈ (-(77.78) - √6180) / (-32.2)
Solving these equations will give us the two possible times when the ball is in the air.