# physics

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An athlete executing a long jump, leaves the ground at a 30 degree angle with the ground and travels 8.90 meters horizontally in 1.1 seconds. What was the takeoff speed of the athlete along the diagonal?

• physics -

D = Xo*t = 8.90 m
Xo * 1.1 = 8.9
Xo = 8.9/1.1 = 8.09 m/s = Hor. component
of initial velocity.

Vo = Xo/cos A = 8.09/cos30=9.34 m/s[30o]
= Initial velocity or take-off speed.

• physics -

km,

• physics -

Δy = 0, assuming cannonball falls back down to same level
a = -g
vi = vsinθ

Δy = vi(t) + ½a(t)²
0 = v(sinθ)(t) + ½(-g)(t)²
0 = t(vsinθ - ½gt)

Clearly, there are two points in time when the projectile is at ground level.
t = 0, is the time when the projectile is launched.
let 0 = vsinθ - ½gt to find out when it lands

½gt = vsinθ
t = 2vsinθ/g

This result can also be obtained by showing that the initial vertical speed is equal to the final vertical speed but opposite in direction:
vf² = vi² + 2gΔy
since change in height is 0,
vf² = vi²
vf = -vi, since the projectile moves up on the way up and down on the way down

Using: a = (vf - vi)/t
t = (vf - vi)/a
t = (-vi - vi)/a
t = -2vi/a
t = -2vsinθ/-g
t = 2vsinθ/g

[Range]

To determine range, analyse motion in the horizontal or x-direction:
Δx = vt, where v is the constant horizontal speed of the projectile
Δx = (vcosθ)(2vsinθ/g)
Δx = v²2cosθsinθ/g

Using the identity: sin(2θ) = 2cosθsinθ,

Δx = v²sin(2θ)/g

v is launch speed
θ is launch angle
g is acceleration due to gravity (9.8 m/s²)

[Launch Speed]

v = sqrt(Δxg/sin(2θ))

Multiply by 1.06 and use new speed in distance equation