a long jump athlete take off at 25 degree with the horizontal and achieves a jumping distance of 9.12 m. Calculate initial take off speed.
To calculate the initial takeoff speed of the long jump athlete, we need to use the following equation:
Horizontal displacement = initial speed * time * cosine(angle)
In this case, the horizontal displacement is given as 9.12 meters, the takeoff angle is 25 degrees, and we need to find the initial speed.
The first step is to convert the angle from degrees to radians, as the trigonometric functions in mathematics operate in radians. To do this, we use the following formula:
Angle in radians = Angle in degrees * (pi / 180)
Plugging in the values:
Angle in radians = 25 * (pi / 180) = 0.4363 radians
Next, we rearrange the equation to solve for the initial speed:
Initial speed = horizontal displacement / (time * cosine(angle in radians))
Since the time is not given in the question, we need more information to solve for the initial speed.
To calculate the initial take-off speed of the long jump athlete, we can use the principles of projectile motion. The given information includes the take-off angle and the jumping distance.
First, let's decompose the initial velocity into its horizontal and vertical components based on the given take-off angle. The horizontal component remains constant throughout the flight, while the vertical component is affected by gravity.
The horizontal component of the initial velocity (Vh) can be calculated using the following formula:
Vh = V * cos(θ)
where V represents the initial take-off speed and θ is the take-off angle (25 degrees).
Next, we can calculate the vertical component of the initial velocity (Vv). The formula for this is:
Vv = V * sin(θ)
The athlete achieves a jumping distance of 9.12 m, which is the horizontal distance covered. Since the horizontal velocity remains constant, we can use the horizontal component of the initial velocity (Vh) and the time of flight (t) to find the take-off speed (V).
The time of flight can be calculated by:
t = 2 * T
where T is the total time taken to reach the maximum height (h).
To find T, we can use the vertical component of the initial velocity (Vv) and gravity (g):
T = Vv / g
Finally, we can calculate the take-off speed (V) using the equation:
V = Vh / t
Substituting the formulas and known values, we can now solve for V.
Vh = V * cos(θ)
Vv = V * sin(θ)
t = 2 * (Vv / g)
V = Vh / t
Now, let's calculate the initial take-off speed:
Vh = V * cos(25°)
Vv = V * sin(25°)
t = 2 * (Vv / g)
V = Vh / t
Substituting the known values:
Vh = V * cos(25°)
Vv = V * sin(25°)
t = 2 * (Vv / 9.8)
V = Vh / t
Now plug in the values:
Vh = V * cos(25°)
Vv = V * sin(25°)
t = 2 * (Vv / 9.8)
V = Vh / t
Vh = V * cos(25°)
Vv = V * sin(25°)
t = 2 * (Vv / 9.8)
V = Vh / t
Solving these equations simultaneously will give us the initial take-off speed (V).
10.8
distanchoriz=vh*timeinair
but time in air can be found by
hf=hi+vy*time-4.9t^2
or 0=0+vy*t-4.9t^2
0=t(vy-4.9) or time in air=vy/4.9
but vy=vsin25 and vh=vcos25
time in air=V*sin25/4.9
9.12=Vcos25*Vsin25/4.9-4.9(V *sin 25/4.9)^2
and you solve with V with some algebra and trig.