A long jumper jumps at a 20 degree angle and attains a maximum altitude of .6 m.
a)what is her initial speed?
b)How far is her jump?
To find the answers to the given questions, we need to break down the problem into its components and make use of relevant physics equations and principles. Let's go step by step.
a) To find the initial speed of the long jumper, we can use the fact that the maximum altitude occurs at the peak of the jump, where the vertical velocity is momentarily zero. At this point, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Here:
vf = final velocity (which is zero at the peak)
vi = initial velocity (what we want to find)
a = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
d = displacement (maximum altitude = 0.6 m)
Plugging in these values into the equation, we get:
0^2 = vi^2 + 2(-9.8)(0.6)
Simplifying the equation, we have:
0 = vi^2 - 11.76
Solving for vi, we take the square root of both sides:
vi = √11.76
Using a calculator, we find that vi ≈ 3.43 m/s.
Therefore, the initial speed of the long jumper is approximately 3.43 m/s.
b) To find the distance of the jump, we can use the horizontal component of the initial velocity. We assume there is no air resistance and neglect any horizontal forces acting on the jumper during the jump. Thus, the horizontal velocity remains constant.
The horizontal distance (range) can be calculated using the equation:
range = horizontal velocity × time of flight
To calculate the horizontal velocity, we can use the initial speed and the angle of 20 degrees. The horizontal component (Vx) is given by:
Vx = vi × cos(θ)
Here:
vi = initial speed (3.43 m/s)
θ = angle of the jump (20 degrees)
Plugging in the values, we have:
Vx = 3.43 × cos(20)
Using a calculator, we find that Vx ≈ 3.225 m/s.
Next, we need to find the time of flight. The time it takes for the long jumper to stay in the air can be calculated using the following equation:
time of flight = (2 × vertical velocity)/g
Here:
vertical velocity = vi × sin(θ) (since there is no initial vertical velocity, we use the value of vi)
g = acceleration due to gravity (9.8 m/s^2)
Plugging in the values, we have:
time of flight = (2 × 3.43 × sin(20))/(-9.8)
Using a calculator, we find that the time of flight ≈ 0.694 seconds.
Finally, we can calculate the horizontal distance (range) using the equation:
range = Vx × time of flight
Plugging in the values, we have:
range = 3.225 × 0.694
Using a calculator, we find that the horizontal distance (range) of the jump is approximately 2.24 meters.
Therefore, the long jumper's jump is approximately 2.24 meters in distance.