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
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?
3 answers
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.
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