An athlete executing a long jump leaves the ground at a 28.9 ∘ angle and travels 7.70 m .What was the takeoff speed?If this speed were increased by just 4.0%, how much longer would the jump be?
VERTICAL SPEED:vi*sin28.9
time in air:
vi=vf-4.9t
but vf=-vi so
t=2vi/4.9=2*vi*sin28.9/4.9
horizontal disance=vicos28.9 *2vi sin28.9/4.9
vi^2=7.7*4.9/2sin28.9*cos28.9
solve for vi.
To find the takeoff speed of the athlete, we can use the following equation:
Horizontal distance = (initial velocity) * (time in the air) * cos(theta)
In this case, the horizontal distance is given as 7.70 meters, and the takeoff angle is given as 28.9 degrees. We need to find the initial velocity.
First, we need to find the time in the air. To do this, we can use the following equation:
Vertical distance = (initial velocity) * (time in the air) * sin(theta) - (1/2) * g * (time in the air)^2
The vertical distance is zero at the highest point of the jump, so we can solve the equation for time in the air:
0 = (initial velocity) * (time in the air) * sin(theta) - (1/2) * g * (time in the air)^2
Rearranging the equation, we get:
(time in the air) = (initial velocity) * sin(theta) / g
Substituting the values given:
(time in the air) = (initial velocity) * sin(28.9) / 9.8
Now we can substitute this value of time in the original equation for horizontal distance:
7.70 meters = (initial velocity) * (initial velocity) * sin(28.9) * cos(28.9) / 9.8
Simplifying the equation, we get:
(initial velocity)^2 = (7.70 meters * 9.8) / (sin(28.9) * cos(28.9))
Taking the square root of both sides, we can find the initial velocity.
Next, to find how much longer the jump would be if the speed increased by 4.0%, we can calculate the new horizontal distance using the increased speed and the same angle:
New horizontal distance = (increased velocity) * (time in the air) * cos(theta)
Then, we can find the difference between the new and original horizontal distances to get the increase in the jump length.
Let's calculate the values step by step.