can somebody show me how to find the time s step by step. A golfer hits his approach shot at an angle of
θ = 46.3°,
giving the ball an initial speed of
v0 = 39.6 m/s
(see figure below). The ball lands on the elevated green,
yf = 4.1 m
above the initial position near the hole, and stops immediately.
How much time passed while the ball was in the air?
s
The equation of ball's height is
y = h0 + (v sinθ)t - 4.9t^2
y = 28.63t - 4.9t^2
So, just solve for y = 4.1
You will get two answers, one for when the ball is going up, and another for when it lands on the green.
steve are you mean take the quadratic of this -4.9t^2+28.63t-4.1
your syntax is strange, but yes, the quadratic formula is the best way to find the desired time.
Vo = 39.6m/s[46.3o].
Xo = 39.6*cos46.3 = 27.4 m/s.
Yo = 39.6*sin46.3 = 28.6 m/s.
Y = Yo + g*Tr = 0 @ max ht.
28.6 + (-9.8)Tr = 0,
Tr = 2.92 s. = Rise time.
Y^2 = Yo^2 + 2g*h = 0.
28.6^2 + (-19.6)*h = 0,
h = 41.7 m. Above launching level.
ho - 0.5g*Tf^2 = 4.1.
41.7 - 4.9Tf^2 = 4.1,
Tf = 2.77 s. = Fall Time.
T = Tr+Tf = 2.92 + 2.77 = 5.69 s. = Time in air.
To find the time it takes for the ball to be in the air, we can use the equations of motion in the x and y directions.
Step 1: Analyze the problem
- The ball is hit at an angle of θ = 46.3°.
- The initial speed of the ball is v0 = 39.6 m/s.
- The ball lands at a vertical displacement yf = 4.1 m.
- The ball stops immediately upon landing.
Step 2: Break down the problem into x and y components
Since we have an angled shot, we need to analyze the motion in both the x and y directions separately.
In the x direction:
- The initial velocity in the x direction (Vx) can be found using the formula Vx = v0 * cos(θ).
- The displacement in the x direction (Sx) can be found using the formula Sx = Vx * t.
In the y direction:
- The initial velocity in the y direction (Vy) can be found using the formula Vy = v0 * sin(θ).
- The displacement in the y direction (Sy) can be found using the formula Sy = Vy * t - 0.5 * g * t^2, where g is the acceleration due to gravity (-9.8 m/s^2)
- The final displacement in the y direction is given as yf = 4.1 m.
Step 3: Solve for time (t)
We need to find the time when the ball lands, which is the time it takes for the displacement in the y direction to reach yf.
Using the equation Sy = Vy * t - 0.5 * g * t^2, substitute in the known values:
4.1 = (v0 * sin(θ)) * t - 0.5 * g * t^2
Since this is a quadratic equation, we can rearrange it and solve for t using the quadratic formula:
0.5 * g * t^2 - (v0 * sin(θ)) * t + yf = 0
Now you can substitute in the known values and solve for t.
Step 4: Calculate the time
Using the values given:
- v0 = 39.6 m/s
- θ = 46.3°
- g = -9.8 m/s^2
- yf = 4.1 m
Plug these values into the quadratic formula and solve for t.
Once you have solved for t, you will have the time it takes for the ball to be in the air.