A short-putter throws the shot with an initial speed of 16 m/s at a 32 degree angle to the horizontal.
Question: Calculate the horizontal distance traveled by the shot if it leaves the athletes hand at a height of 2.05 m above the ground?
- 45.8m was not the correct answer and I have worked the problem several times over
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•t1²/2 =16•1.63-9.8•(1.63)²/2 =13,1 m.
L1= vₒ•cos32•t1=16•0.85•1.63=22,12 m.
H=h+hₒ=13,1 + 2.05=15.15 m.
H=g•t2²/2.
t2=sqrt(2H/g) = sqrt(2•15.15/9.8) =1.76 s.
L2= vₒ•cos32•t2=16•0.85•1.76=23.94 m.
L =L1+L2 = 22.12+23.94 = 46.06 m.
To calculate the horizontal distance traveled by the shot, we can use the kinematic equations of motion. The horizontal distance can be found by determining the time of flight and then multiplying it by the horizontal component of the initial velocity.
First, let's determine the time of flight. The initial vertical velocity of the shot can be found using trigonometry:
Vy = V * sin(theta)
Where Vy is the vertical component of the initial velocity, V is the initial speed (16 m/s), and theta is the launch angle (32 degrees).
Vy = 16 m/s * sin(32 degrees)
Vy ≈ 8.527 m/s
To find the time of flight, we can use the equation:
t = (2 * Vy) / g
Where t is the time of flight and g is the acceleration due to gravity (approximately 9.8 m/s^2).
t = (2 * 8.527 m/s) / 9.8 m/s^2
t ≈ 1.74 s
Now that we know the time of flight, we can find the horizontal distance traveled. The horizontal component of the initial velocity can be found by:
Vx = V * cos(theta)
Where Vx is the horizontal component of the initial velocity, V is the initial speed (16 m/s), and theta is the launch angle (32 degrees).
Vx = 16 m/s * cos(32 degrees)
Vx ≈ 13.555 m/s
The horizontal distance can be calculated using the formula:
d = Vx * t
Where d is the horizontal distance traveled.
d = 13.555 m/s * 1.74 s
d ≈ 23.59 m
Therefore, the horizontal distance traveled by the shot is approximately 23.59 meters.