Athelete jump finding starting speed

[mortho]

An athlete executing a long jump leaves the ground at a 40° angle and travels 6.30 m.

a) what was the take off speed?
b) If the speed was increased by just 4.0 percent, how much longer would the jump be?

** i used squareroot of 2ax to find initial velocity and got 8.91 m/s but got it wrong. what was the problem>?

 

[dynamicsolo]

An athlete executing a long jump leaves the ground at a 40° angle and travels 6.30 m.

a) what was the take off speed?
b) If the speed was increased by just 4.0 percent, how much longer would the jump be?

** i used squareroot of 2ax to find initial velocity and got 8.91 m/s but got it wrong. what was the problem>?

Since the jumper is traveling in two dimensions (vertically as well as horizontally), you cannot simply use the "velocity-squared" formula to find their starting speed.

The jumper takes on with an unknown speed v0 at a 40º angle to the horizontal. What does that mean for their starting horizontal and vertical velocities? How long will the jumper stay in the air before landing? You are told how far they moved horizontally before touching down.

Once you have part (a), that will give you an idea of how to deal with part (b).

 

[ogb p]

I can provide an explanation for the incorrect answer that was obtained using the formula √(2ax) to calculate the initial velocity. This formula is generally used for calculating the initial velocity in a projectile motion scenario where the object is launched at an angle and lands at the same height as it started. In the case of the athlete's long jump, the height at takeoff and landing are not the same, so this formula cannot be used.

To accurately calculate the takeoff speed in this scenario, we need to use the formula v = √(g * d/cosθ), where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), d is the horizontal distance traveled (6.30 m), and θ is the angle of takeoff (40°). Plugging in these values, we get v = √(9.8 * 6.30/cos40°) = 9.61 m/s.

To answer the second part of the question, if the speed is increased by 4%, the new initial velocity would be 9.61 * 1.04 = 9.98 m/s. Using the same formula, we can calculate the new distance traveled as d = v²*sin2θ/g. Plugging in the new values, we get d = (9.98)²*sin80°/9.8 = 6.54 m. This means that increasing the speed by 4% would result in an increase in the jump distance by 0.24 m.

In conclusion, the incorrect answer was obtained due to the use of an inappropriate formula for the given scenario. By using the correct formula, we can accurately calculate the takeoff speed and the potential increase in jump distance with a 4% increase in speed.