An athlete executing a long jump leaves the ground at a 32.5° angle and travels 7.72 m. What was the take-off speed?

If this speed were increased by just 4.9 %, how much longer would the jump be?

The formula you need to answer both questions is

Distance = Vo^2*sin(2A)/g

where A it the takeoff angle.

Deriving that formula would be a good exercise for you. You would use the trig identity sin(2A) = 2 sinA cosA.

In the first case,
Vo^2 = g*7.72m/sin65 = 83.56 m^2/s^2
Take the square root of that for Vo.

Increasing the speed by a factor 1.049 will increase the length of the jump by a factor (1.049)^2 = 1.100
(a 10.0% increase).

To find the take-off speed of the athlete executing the long jump, we can use the equations of motion. The vertical motion and horizontal motion can be considered independently.

First, let's find the initial vertical velocity (Vy) component of the jump. We can use the equation:

Vy = V0 * sin(theta)

Where Vy is the vertical velocity component, V0 is the initial velocity (take-off speed), and theta is the angle of take-off which is 32.5° in this case.

Vy = V0 * sin(32.5°)

To find the initial horizontal velocity (Vx) component of the jump, we can use the equation:

Vx = V0 * cos(theta)

Where Vx is the horizontal velocity component.

Vx = V0 * cos(32.5°)

Since the time of flight is the same for both horizontal and vertical components, we can use the equation:

t = d / Vx

Where t is the time of flight, and d is the horizontal distance traveled which is 7.72 m in this case.

t = 7.72 m / (V0 * cos(32.5°))

Now, let's find the time of flight (t) using the vertical motion equation:

Vy = V0 * sin(theta) - g * t

Where g is the acceleration due to gravity which is approximately 9.8 m/s².

0 = V0 * sin(32.5°) - 9.8 m/s² * t

Rearranging the equation to isolate t:

t = V0 * sin(32.5°) / 9.8 m/s²

Now we can substitute this expression for t into the equation for Vx:

Vx = V0 * cos(32.5°)

t = 7.72 m / (V0 * cos(32.5°))

Combining the two expressions for t, we get:

V0 * sin(32.5°) / 9.8 m/s² = 7.72 m / (V0 * cos(32.5°))

Now we can solve for V0 by cross-multiplying and rearranging the equation:

V0^2 * sin(32.5°) * cos(32.5°) = 7.72 m * 9.8 m/s²

V0^2 = (7.72 m * 9.8 m/s²) / (sin(32.5°) * cos(32.5°))

V0^2 = 90.9732 m²/s²

Taking the square root of both sides:

V0 = sqrt(90.9732 m²/s²)

V0 ≈ 9.54 m/s

Therefore, the take-off speed of the athlete is approximately 9.54 m/s.

Now, to find the increase in the jump length when the speed is increased by 4.9%:

Let's assume the new speed is V_new, and the jump length is d_new.

d_new = V_new * t

The new speed is increased by 4.9%, which means the increase is:

Increase_in_speed = 4.9% of V0

Increase_in_speed = (4.9/100) * V0

d_new = (V0 + Increase_in_speed) * t

To find the increase in jump length, we need to determine the difference between the new jump length (d_new) and the original jump length (d).

Increase_in_jump = d_new - d

Substituting the values we have:

Increase_in_jump = ((V0 + Increase_in_speed) * t) - (V0 * t)

Simplifying:

Increase_in_jump = (V0 + Increase_in_speed - V0) * t

Increase_in_jump = Increase_in_speed * t

Substituting the values of Increase_in_speed and t:

Increase_in_jump = [(4.9/100) * V0] * t

Calculating:

Increase_in_jump = [(4.9/100) * 9.54 m/s] * (7.72 m / (9.54 m/s * cos(32.5°)))

Increase_in_jump ≈ (0.049 * 9.54 m/s) * (7.72 m / (9.54 m/s * cos(32.5°)))

Therefore, the increase of the jump length when the speed is increased by 4.9% is approximately equal to (0.049 * 9.54 m/s) * (7.72 m / (9.54 m/s * cos(32.5°))).