An athlete executing a long jump leaves the ground at a 30.3 angle and travels 7.68 .

What is the take off speed?

If this speed were increased by just 3.0%, how much longer would the jump be?

To find the takeoff speed, we can use the principles of projectile motion.

First, let's break down the information given:
- Angle of takeoff (θ) = 30.3°
- Distance traveled (d) = 7.68 m

We can use the following equation to determine the takeoff speed (v):
d = (v^2 * sin(2θ)) / g

Here, g represents the gravitational acceleration, which is approximately 9.8 m/s^2.

Rearranging the equation, we have:
v = sqrt((d * g) / sin(2θ))

Now, let's substitute the values given into the equation and calculate the takeoff speed:

θ = 30.3°
d = 7.68 m
g = 9.8 m/s^2

v = sqrt((7.68 * 9.8) / sin(2*30.3))

Calculating the expression inside the square root:
sin(2*30.3) ≈ 0.9876

v = sqrt((7.68 * 9.8) / 0.9876)
v ≈ sqrt(75.398/0.9876)
v ≈ sqrt(76.328)

Therefore, the takeoff speed is approximately 8.74 m/s.

Now, let's move on to the second part of the question.

If the takeoff speed is increased by 3.0%, we need to find the new distance traveled.

To do this, we can calculate the new takeoff speed by adding 3.0% of the original speed to the original speed. Then, we can calculate the new distance using the same equation.

First, let's calculate the new takeoff speed:
New takeoff speed = 1.03 * 8.74
New takeoff speed ≈ 9.00 m/s (rounded to two decimal places)

Now, we can plug the new takeoff speed into the equation to calculate the new distance:
New distance traveled = (v^2 * sin(2θ)) / g

Using the same values for θ and g, and substituting the new takeoff speed, we have:
New distance traveled = (9.00^2 * sin(2*30.3)) / 9.8

Calculating the expression inside the square root:
sin(2*30.3) ≈ 0.9876

New distance traveled = (81.00 * 0.9876) / 9.8
New distance traveled ≈ 8.15 m (rounded to two decimal places)

Therefore, increasing the takeoff speed by 3.0% will result in the jump being approximately 0.47 m longer.