A long jump athlete take off at 25 degree with the horizo and achieves a jumping distance of 9.12 m.
(i) calculate the initial take-off speed
(ii) calculate the maximum height of the jump.
R=9.12m and @=25.
Its initial velocity is by
R=Vo^2.sin2@/g
Vo=10.8 m/s
Maximum height is achieved at mid point of distance covered, then
tan@ = max height/(R/2)
h=2.12 m
To calculate the initial take-off speed, we can use the following steps:
Step 1: Convert the take-off angle from degrees to radians.
The angle of 25 degrees needs to be converted to radians by multiplying it by (π/180):
θ = 25° × (π/180) = 0.4363 radians (approx.)
Step 2: Use the range formula to calculate the initial take-off speed.
The range formula is given by:
R = (v^2 * sin(2θ)) / g
where:
R = jumping distance = 9.12 m
v = initial take-off speed (what we want to find)
θ = take-off angle (0.4363 radians)
g = acceleration due to gravity = 9.8 m/s^2
Now, we can rearrange the formula to solve for v:
v^2 = (R * g) / sin(2θ)
v = √((R * g) / sin(2θ))
Substituting the values, we get:
v = √((9.12 * 9.8) / sin(2 * 0.4363))
Calculating the expression, we find:
v ≈ 8.05 m/s (approx.)
Therefore, the initial take-off speed is approximately 8.05 m/s.
To calculate the maximum height of the jump, we can use the following steps:
Step 1: Use the formula for the maximum height of a projectile:
h_max = (v^2 * (sin(θ))^2) / (2 * g)
where:
h_max = maximum height (what we want to find)
v = initial take-off speed (8.05 m/s)
θ = take-off angle (0.4363 radians)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the values, we get:
h_max = (8.05^2 * (sin(0.4363))^2) / (2 * 9.8)
Calculating the expression, we find:
h_max ≈ 0.638 m (approx.)
Therefore, the maximum height of the jump is approximately 0.638 m.
To calculate the initial take-off speed of the long jump athlete, we can use the following equations:
(i) Calculate the horizontal component of initial velocity:
The horizontal component of the initial velocity (Vx) can be calculated using the formula:
Vx = V * cos(theta)
where:
Vx is the horizontal component of the velocity,
V is the initial take-off speed, and
theta is the take-off angle with respect to the horizontal.
Given:
theta = 25 degrees,
Vx = V * cos(25)
(ii) Calculate the vertical component of initial velocity:
The vertical component of the initial velocity (Vy) can be calculated using the formula:
Vy = V * sin(theta)
where:
Vy is the vertical component of the velocity.
Given:
theta = 25 degrees,
Vy = V * sin(25)
With the given information, we can proceed to solve the problem step-by-step:
(i) To calculate the initial take-off speed (V), we need to find the value of Vx:
Vx = V * cos(25)
Now we rearrange the equation to solve for V:
V = Vx / cos(25)
Substituting the given values:
V = 9.12 / cos(25)
Using a calculator, evaluate cos(25):
V = 9.12 / 0.9063
V ≈ 10.06 m/s (rounded to two decimal places)
Therefore, the initial take-off speed of the long jump athlete is approximately 10.06 m/s.
(ii) To calculate the maximum height of the jump, we can use the following formula:
The maximum height (H) can be calculated using the formula:
H = (Vy^2) / (2 * g)
where:
H is the maximum height,
Vy is the vertical component of the velocity, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
Vy = V * sin(25)
Now we rearrange the equation to solve for H:
H = (V * sin(25))^2 / (2 * g)
Substituting the given values:
H = (10.06 * sin(25))^2 / (2 * 9.8)
Using a calculator, evaluate sin(25), then evaluate the equation:
H ≈ 1.84 m (rounded to two decimal places)
Therefore, the maximum height of the jump is approximately 1.84 meters.