An athlete performing a long jump leaves the ground at a 27.1∘ angle and lands 7.72m away
1.What was the takeoff speed?
2.If this speed were increased by just 5.0%, how much longer would the jump be?
u = speed * cos 27.1
7.72 = speed *cos 27.1 * t
where t is time in air
h = Hi + Vi t - (1/2) g t^2
0 = 0 + speed*sin 27.1 * t - 4.9 t^2
but we know t = 7.72/(speed*cos 27.1)
call speed s
0 = 0 + 7.72 tan 27.1 - 4.9*7.72^2/(s^2cos^227.1)
solve for s
do it again for s = 1.05 s
To find the takeoff speed of the athlete performing a long jump, we can use the following kinematic equation:
Range = (v^2 * sin(2θ)) / g
Where:
- Range is the horizontal distance covered in the long jump (7.72m in this case)
- v is the takeoff speed we want to find
- θ is the takeoff angle (27.1 degrees in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
1. Finding the takeoff speed:
To find the takeoff speed (v), we rearrange the equation:
v = sqrt((Range * g) / sin(2θ))
Plugging in the values:
v = sqrt((7.72 * 9.8) / sin(2 * 27.1))
Calculating this expression will give us the takeoff speed.
2. Finding the increase in jump length with a 5.0% increase in speed:
To calculate the increase in jump length, we need to find the new range when the speed is increased by 5.0%. Let's call the new speed v_new.
v_new = 1.05 * v
Using the same kinematic equation as before, we can substitute v_new into the equation instead of v to find the new range:
New Range = (v_new^2 * sin(2θ)) / g
Substituting the value of v_new:
New Range = ((1.05 * v)^2 * sin(2θ)) / g
Calculating this expression will give us the new range.
To find the difference in jump length, we subtract the original range from the new range:
Difference in Jump Length = New Range - Range
Calculating this expression will give us the increase in jump length with a 5.0% increase in speed.