An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7 meters before landing. What is the takeoff speed of the jumper?
Well this must be a pretty common question because I had it too. The problem here is the Voy is Vo(SIN(23) not cos)/10.0 (or 9.8).
Also because the Vfy is 0 in the middle of the jump used only use half the X for the horizontal component. This is how I got the 11 m/s initial take off speed.
I don't know for sure, but that's how I'm solving these type of questions.
Physics question answer
To find the takeoff speed of the Olympic long jumper, we need to apply conservation of energy and kinematic equations. The key principle we're going to use is that the initial kinetic energy of the jumper is equal to the final potential energy (when the jumper lands). Here's how we can calculate it:
1. Start by considering the vertical motion of the jumper. We can use the kinematic equation for vertical motion:
vf² = vi² + 2gh
where vf is the final vertical velocity (0 m/s when the jumper lands), vi is the initial vertical velocity (which we need to find), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the maximum height reached by the jumper.
2. The maximum height reached by the jumper is determined by the vertical component of the takeoff velocity (viy):
viy = vi * sin(θ)
where θ is the takeoff angle (23 degrees in this case).
3. The horizontal distance traveled by the jumper (8.7 meters) can be calculated using the horizontal component of the takeoff velocity (vix):
vix = vi * cos(θ)
Using this, we can find the time of flight (t) using the formula:
d = vix * t
4. Now that we have the time of flight, we can calculate the maximum height using the formula:
h = viy * t + (1/2) * g * t²
5. Plugging the value of h back into the first equation, we can solve for vi:
vf² = vi² + 2gh, where vf = 0
6. Solve for vi to find the takeoff speed of the jumper.
Applying these steps, we can find the takeoff speed of the jumper.
Note: It's important to convert the angle from degrees to radians when using the trigonometric functions, as they typically take radians as an input.
Let the initial velocity = Vo
The horizontal comonent, Vx = Vo(cos23)
The horizontal distance, x = Vo(cos23)t = 8.7m
t = 8.7m / Vo(cos23)
The vertical component of initial velocity is Vyo:
Vyo = Vo(sin23)
Let the final vertical velocity be Vyf. Then,
Vyf = 0 (a the highest position)
0 = Vo(cos23) - 9.8m/s^2*t, or:
t = Vo(cos23) / 9.8m/s^2
You have two expressions for time, t.
Since the two are equal, you can solve for Vo