an athlete executing a long jump leaving the ground at an angle of 30° and travels 8.9m .what was the take off speed?
time in air:
distance=v*cos30*t
t= 8.9/v*.866
take off speed:
v*sin30 is vertical velocity (vi')
hf=hi+vi'*t-4.8t^2 solve for v, take off speed, using time in air t as above.
0=0=v*sin30*(8.9/v*.866)-4.8(8.9/v*.866)^2
or v^2*.5*8.9/.866=4.8*(8.9/.866)^2
v^2=4.8*8.9/.866 check my math, take square root
To find the takeoff speed of the athlete, we can use the principles of projectile motion and break down the motion into horizontal and vertical components.
We know that the horizontal distance traveled (range) is 8.9 meters. However, since there is no acceleration in the horizontal direction, the horizontal component of the speed remains constant throughout the motion. Therefore, the horizontal component of the speed can be calculated as:
Horizontal speed = Range / Time
Keep in mind that the time of flight can be found using the vertical motion of the athlete.
Using the vertical motion, we can determine the time of flight considering the equation:
h = ut + (1/2)gt^2
Where:
h = vertical displacement (in this case, h = 0 since the athlete returns to the ground)
u = initial vertical velocity (takeoff speed)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of flight
Since the vertical displacement is zero, the equation simplifies to:
0 = ut - (1/2)gt^2
We can solve this quadratic equation for t. In this case, there will be two roots for t, one of which will be negative (meaning that the athlete has not yet taken off) and the other will be positive (representing the time of flight).
Once we have the time of flight, we can substitute it back into the horizontal speed equation to find the takeoff speed.
Let's calculate the solution step by step:
Step 1: Calculate the time of flight (t)
Using the quadratic equation ut - (1/2)gt^2 = 0, we can find the time of flight. Since the initial velocity is unknown, we will use a variable v:
v * t - (1/2)g * t^2 = 0
Simplifying the equation:
t * (v - (1/2)g * t) = 0
This equation will hold true for two scenarios, either t = 0 or (v - (1/2)g * t) = 0.
Setting (v - (1/2)g * t) = 0:
v - (1/2)g * t = 0
Substituting g = 9.8 m/s^2:
v - (1/2) * 9.8 * t = 0
v - 4.9t = 0
v = 4.9t
Since we know the angle at takeoff is 30°, we can determine the vertical speed component using trigonometry.
The vertical component of speed (v_y) is given by:
v_y = v * sin(θ)
Where:
v_y = vertical component of speed
v = takeoff speed
θ = angle of takeoff (30°)
Therefore:
v_y = v * sin(30°)
v_y = (v * √3) / 2
Now, we can rewrite the equation as:
(v * √3) / 2 - 4.9t = 0
Step 2: Calculate the time of flight (t)
Rearranging the equation:
(v * √3) / 2 = 4.9t
t = (v * √3) / (2 * 4.9)
t = (v * √3) / 9.8
Step 3: Calculate the horizontal speed
Now that we have the time of flight, we can calculate the horizontal speed:
Horizontal speed = Range / Time
Horizontal speed = 8.9 / t
Substituting t = (v * √3) / 9.8:
Horizontal speed = 8.9 / ((v * √3) / 9.8)
Simplifying the expression:
Horizontal speed = (1.96 * 8.9) / v
Step 4: Solve for the takeoff speed (v)
Since the horizontal speed is equal to the horizontal component of the velocity, we can equate the two:
Horizontal speed = v * cos(θ)
Where:
θ = angle of takeoff (30°)
Therefore:
(1.96 * 8.9) / v = v * cos(30°)
Now, solve this equation to find the value of v:
(1.96 * 8.9) = v^2 * cos(30°)
v^2 = (1.96 * 8.9) / cos(30°)
v = √((1.96 * 8.9) / cos(30°))
v ≈ 5.03 m/s
So, the takeoff speed of the athlete is approximately 5.03 m/s.