Question

Calculate the initial speed, v0, of a long-jumper where d=7.56 m. Assume a take-off angle of 22.5°, an initial height of the centre-of-mass of 0.920 m, and a final height of 0.460 m.

Answer 1

Here's a visual diagram

Answer 2

try setting up two equaitions and solve :-
1) equations for vertical position
2) equaton for horizontal position

Answer 3

you may take that point as origin

Answer 4

can u setup an equaiton for vertical position ?

Answer 5

Equation for height (y):\[y=h_0+v_0\sin(\theta)·t-\frac{ 1 }{ 2 }g·t^2\]Equation for distance(x)\[x=d_0+v_0\cos(\theta)·t \rightarrow t=\frac{ x-d_0 }{ v_0\cos(\theta) }\]We replace this expression of time in the first equation and get:\[y=h_0+(x-d_0)\tan(\theta)-\frac{ 1 }{ 2}g \left[ \frac{ x-d_0 }{ v_0\cos(\theta) } \right]^2\]Which is the equation of a parabola. This function has to touch the landing point too:\[h_1=h_0+(d_1-d_0)\tan(\theta)-\frac{ 1 }{ 2 }g \left[\frac{ d_1-d_0 }{ v_0\cos(\theta) }\right]^2\]This an expression that relates: Initial height (h0), final height(h1),initial distance(d0), final distance(d1), initial velocity (Vo) and angle (θ).
In your problem, you have:
d0=0
d1=7.56
h0=0.920
h1=0.460
θ=22.5º
That means enough data to calculate Vo