A shot-putter with an initial speed of 16m/s at 32 degree angle to the horizontal.

Calculate the horizontal distance traveled by the shot if it leaves the athletes hand at a height of 2.05m above the ground?

To calculate the horizontal distance traveled by the shot, we need to break down the initial velocity into its horizontal and vertical components.

Given:
- Initial speed, v0 = 16 m/s
- Launch angle, θ = 32 degrees
- Initial height, h = 2.05 m

First, let's calculate the vertical component of the initial velocity (Vy) using the equation:
Vy = v0 * sin(θ)
Vy = 16 m/s * sin(32 degrees)
Vy ≈ 8.51 m/s

Next, let's calculate the time of flight (t) using the equation for vertical motion:
h = (1/2) * g * t^2
2 * h / g = t^2
t = sqrt(2 * h / g)
where g is the acceleration due to gravity (9.8 m/s^2)
t = sqrt(2 * 2.05 m / (9.8 m/s^2))
t ≈ 0.6422 seconds

Now, let's calculate the horizontal component of the initial velocity (Vx) using the equation:
Vx = v0 * cos(θ)
Vx = 16 m/s * cos(32 degrees)
Vx ≈ 13.66 m/s

Finally, we can calculate the horizontal distance traveled (d) using the equation:
d = Vx * t
d ≈ 13.66 m/s * 0.6422 seconds
d ≈ 8.77 meters

Therefore, the horizontal distance traveled by the shot is approximately 8.77 meters.

Upwards motion

v=vₒ-g•t1.
At the top point
0= vₒ-g•t1,
t1= vₒ/g=16/9.8=1.63.
h=vₒ•t1-g•t1²/=16•1.63-9.8•(1.63)²/2 =13 m.
L1= vₒ•cos32•t1=16•0.85•1.63=22 m.
H=h+hₒ=13+2.05=15.05 m.
H=g•t2²/2.
t2=sqrt(2H/g) = sqrt(2•15.05/9.8) =1.75 s.
L2= vₒ•cos32•t2=16•0.85•1.75=23.8 m.
L =L1+L2 = 22+23.8 = 45.8 m.