An athlete performing a long jump leaves the ground at a 27.1 ∘ angle and lands 7.73 m away.
1). What was the takeoff speed?
Express your answer using three significant figures and include the appropriate units.
2).If this speed were increased by just 3.0%, how much longer would the jump be?
Express your answer using two significant figures and include the appropriate units.
see earlier post
1. Range = Vo^2 * sin(2A)/g.
7.73 = Vo^2 * sin(54.2)/9.8.
7.73 = Vo^2 * 0.0828, Vo^2 = 93.4, Vo = 9.66 m/s.
Range = (9.66*1.03)^2 * sin(54.2)/9.8 =
To answer these questions, we can use the principles of projectile motion. Let's break them down step by step:
1) To find the takeoff speed, we can use the horizontal distance (range) and the launch angle. The horizontal distance is given as 7.73 m, and the launch angle is given as 27.1°. In projectile motion, the horizontal and vertical components of motion are independent of each other. Therefore, we can use the horizontal component of the motion to find the takeoff speed.
The horizontal component of the velocity is given by the formula:
Vx = V * cos(θ)
where V is the initial velocity (takeoff speed) and θ is the launch angle. In this case, we know Vx = 7.73 m and θ = 27.1°.
Rearranging the formula, we have:
V = Vx / cos(θ)
Plugging in the given values, we get:
V = 7.73 m / cos(27.1°)
Calculating this value, we find that the takeoff speed is approximately 8.89 m/s. Therefore, the answer to question 1 is:
Takeoff speed = 8.89 m/s
2) Next, let's consider what happens if we increase the takeoff speed by 3.0%.
Since the increase in speed doesn't affect the launch angle or any other variables, we can assume the trajectory will remain the same. In other words, the athlete will still take off at a 27.1° angle.
To find how much longer the jump would be, we need to calculate the new horizontal distance (range) using the increased takeoff speed.
Let's call the increased takeoff speed V'. Then we have:
V' = V + 0.03V
Simplifying this equation, we get:
V' = 1.03V
Now we can calculate the new horizontal distance by plugging in the values:
R' = (V' * cos(θ)) * t
where R' is the new horizontal distance, V' is the increased takeoff speed, θ is the launch angle, and t is the time of flight.
Since we are comparing the new range to the original range, t will be the same in both cases. Therefore, we can simplify the equation further:
R' = (1.03V * cos(θ)) * t
Dividing the equation by the original range, we get:
(R' - R) / R = (1.03V * cos(θ) * t - V * cos(θ) * t) / (V * cos(θ) * t)
The t cancels out, and we can simplify the equation to:
(R' - R) / R = (1.03V - V) / V
Plugging in the values we know, R = 7.73 m and V = 8.89 m/s, we can calculate the increase in range:
(R' - R) / R = (1.03 * 8.89 - 8.89) / 8.89
Calculating this value, we find that the increase in range is approximately 0.30 m. Therefore, the answer to question 2 is:
Increase in jump length = 0.30 m