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? If this speed were increased by just 5.0%, how much longer would the jump be?
To find the takeoff speed of the athlete, we can use the kinematic equation for horizontal distance:
d = v₀ * t
where:
d is the horizontal distance traveled (7.89 m)
v₀ is the initial velocity (takeoff speed)
t is the time of flight
Now, we need to find the time of flight using the kinematic equation for vertical displacement:
h = v₀ * sin(θ) * t - (1/2) * g * t²
where:
h is the vertical displacement (0 m, since the athlete lands at the same height)
θ is the takeoff angle (27.0 degrees)
g is the acceleration due to gravity (approximately 9.8 m/s²)
Rearranging this equation, we can solve for t:
t = 2 * v₀ * sin(θ) / g
Plugging this value of t into the equation for horizontal distance:
d = v₀ * (2 * v₀ * sin(θ) / g)
Now, we can solve this equation for v₀:
v₀ = √(d * g / (2 * sin(θ)))
Substituting the given values:
θ = 27.0 degrees
d = 7.89 m
g = 9.8 m/s²
v₀ = √(7.89 * 9.8 / (2 * sin(27.0)))
Using a calculator, we can find that v₀ is approximately 8.47 m/s (rounded to two decimal places).
Now let's calculate the increased speed by 5.0%:
increase = 5.0%
increased_speed = v₀ * (1 + increase)
increased_speed = 8.47 m/s * (1 + 5.0%)
increased_speed = 8.47 m/s * (1 + 0.05)
increased_speed = 8.47 m/s * 1.05
increased_speed = 8.895 m/s (rounded to three decimal places)
To find how much longer the jump would be with the increased speed, we can use the same equation for horizontal distance:
increased_distance = increased_speed * t
Substituting the known values of increased_speed:
increased_distance = 8.895 m/s * (2 * 8.895 m/s * sin(27.0)) / 9.8 m/s²
Using a calculator, we can calculate that the increased distance is approximately 9.94 m (rounded to two decimal places).
Therefore, increasing the takeoff speed by 5.0% would make the jump approximately 9.94 m - 7.89 m = 2.05 m longer.