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?

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.

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

To find the takeoff speed of the athlete along the diagonal, we need to break down the initial velocity into horizontal and vertical components.

Given:
Angle with the ground (θ) = 30 degrees
Horizontal distance traveled (d) = 8.90 meters
Time taken (t) = 1.1 seconds

Step 1: Find the horizontal component of the initial velocity.
The horizontal component of the velocity (Vx) can be found using the formula:
Vx = d / t

Substituting the given values, we have:
Vx = 8.90 m / 1.1 s
Vx ≈ 8.09 m/s (rounded to two decimal places)

Step 2: Find the vertical component of the initial velocity.
The vertical component of the velocity (Vy) can be found using the formula:
Vy = V * sin(θ)

Substituting the given angle, we have:
Vy = V * sin(30°)

Step 3: Find the takeoff speed along the diagonal.
The speed along the diagonal (V) can be found using the Pythagorean theorem:
V = √(Vx² + Vy²)

Substituting the values we found:
V = √(8.09² + (V * sin(30°))²)

Squaring both sides and rearranging the equation, we have:
V² = 8.09² + (V * sin(30°))²
V² = 65.68 + V² * (1/2)²
V² = 65.68 + V² * 1/4
3/4 * V² = 65.68
V² = (4/3) * 65.68
V² = 87.57333...

Taking the square root of both sides, we have:
V ≈ √87.57333...
V ≈ 9.36 m/s (rounded to two decimal places)

So, the takeoff speed of the athlete along the diagonal is approximately 9.36 m/s.

To find the takeoff speed of the athlete along the diagonal, we can use the components of the athlete's motion.

Given:
Angle with the ground (θ): 30 degrees
Horizontal distance traveled (dx): 8.90 meters
Time of flight (t): 1.1 seconds

First, let's find the vertical component of the athlete's motion:

We know that the vertical acceleration due to gravity (g) is -9.8 m/s² (negative because it acts downward).

Using the equation for vertical displacement (dy) in projectile motion:
dy = V₀ * sin(θ) * t + (1/2) * g * t²

Since the athlete leaves the ground at a 30 degree angle, the initial vertical velocity component (V₀ * sin(θ)) is zero at takeoff. Therefore, the equation simplifies to:
dy = (1/2) * g * t²

Now, we can calculate the vertical displacement (dy):
dy = (1/2) * (-9.8 m/s²) * (1.1 s)²
dy ≈ -6.36 meters

Next, let's find the horizontal component of the athlete's motion:

Using the equation for horizontal displacement (dx) in projectile motion:
dx = V₀ * cos(θ) * t

Rearranging the equation, we can solve for the initial horizontal velocity component (V₀ * cos(θ)):
V₀ * cos(θ) = dx / t

Now, we can calculate the initial horizontal velocity component (V₀ * cos(θ)):
V₀ * cos(θ) = 8.90 m / 1.1 s
V₀ * cos(30°) ≈ 8.09 m/s

Finally, we can find the total initial velocity (V₀) along the diagonal using the horizontal and vertical components:

Using the Pythagorean theorem:
V₀ = √((V₀ * cos(θ))² + (V₀ * sin(θ))²)

Substituting the values:
V₀ = √((8.09 m/s)² + (0 m/s)²)
V₀ ≈ 8.09 m/s

Therefore, the takeoff speed of the athlete along the diagonal is approximately 8.09 m/s.