An athlete executing a long jumper leaves the ground at a 27.0 degree angle and travels 7.89 m. What was the takeoff speed?
To find the takeoff speed of the athlete executing a long jump, we can use the horizontal and vertical components of the motion.
Given:
Angle of takeoff (θ) = 27.0 degrees
Distance traveled (d) = 7.89 m
Step 1: Break down the motion into horizontal and vertical components.
The horizontal component is given by: Vx = V * cos(θ)
The vertical component is given by: Vy = V * sin(θ)
Step 2: Solve for the vertical component.
We know that the vertical displacement (Δy) for projectile motion can be calculated using the equation:
Δy = (Vy^2) / (2 * g)
Where g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the values:
7.89 m = (Vy^2) / (2 * 9.8 m/s^2)
Step 3: Solve for the vertical component of the velocity (Vy).
Rearranging the equation from step 2, we get:
Vy^2 = 2 * 9.8 m/s^2 * 7.89 m
Vy^2 = 154.392 m^2/s^2
Vy = √(154.392 m^2/s^2)
Vy ≈ 12.42 m/s
Step 4: Solve for the horizontal component of the velocity (Vx).
Using the equation from step 1:
Vx = V * cos(θ)
Vx = V * cos(27.0 degrees)
Step 5: Solve for the takeoff speed (V).
We can use the Pythagorean theorem to find the overall velocity (V) as follows:
V = √(Vx^2 + Vy^2)
V = √((V * cos(27.0 degrees))^2 + (12.42 m/s)^2)
Step 6: Solve for V using algebraic manipulation:
V^2 = (V * cos(27.0 degrees))^2 + (12.42 m/s)^2
V^2 = V^2 * cos^2(27.0 degrees) + 154.392 m^2/s^2
V^2 - V^2 * cos^2(27.0 degrees) = 154.392 m^2/s^2
V^2 * (1 - cos^2(27.0 degrees)) = 154.392 m^2/s^2
V^2 * sin^2(27.0 degrees) = 154.392 m^2/s^2
Step 7: Solve for V:
V = √(154.392 m^2/s^2 / sin^2(27.0 degrees))
V ≈ √(365.411 m^2/s^2)
V ≈ 19.12 m/s
Therefore, the takeoff speed of the athlete is approximately 19.12 m/s.
To find the takeoff speed of the athlete executing a long jump, we can use the equations of horizontal and vertical motion.
We know that the horizontal motion is not affected by gravity, so the horizontal velocity remains constant throughout the jump. Thus, we can focus on the vertical motion.
The vertical motion can be analyzed using the equation of motion:
h = ut + (1/2)gt^2
Where,
h = vertical displacement (7.89 m)
u = initial vertical velocity (takeoff speed)
t = time of flight
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the jump starts and ends at the same height (ground level), the vertical displacement h is 0. Therefore, we can rewrite the equation as:
0 = ut - (1/2)gt^2
Simplifying the equation, we get:
ut = (1/2)gt^2
Now, let's focus on the horizontal motion. The horizontal distance traveled is given as 7.89 m.
We know that the horizontal displacement (7.89 m) is equal to the horizontal velocity (constant) multiplied by the time of flight (t).
Therefore, we can write:
7.89 m = u*cos(27°)*t
Now, we have two equations relating the vertical and horizontal components. We can solve them simultaneously to find the takeoff speed (u).
Dividing the first equation by t and rearranging, we get:
u = (1/2)g*t
Substituting this value of u into the second equation, we have:
7.89 m = [(1/2)g*t]*cos(27°)*t
Simplifying further:
7.89 m = (1/2)*9.8 m/s^2*t^2*cos(27°)
Now, we can solve this equation to find the value of t.
1. Rearrange the equation to isolate the variable:
t^2 = (2*7.89 m) / (0.5*9.8 m/s^2*cos(27°))
2. Calculate the value of t by taking the square root of both sides of the equation.
3. Once you have the value of t, you can substitute it back into the equation for u:
u = (1/2)g*t
This will give you the takeoff speed (u) of the athlete executing the long jump.