A short-putter throws the shot with an initial speed of 16 m/s at a 32.0 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?
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.
To calculate the horizontal distance traveled by the shot, we can use the equations of motion. The horizontal and vertical motions of the shot are independent of each other, so we can treat them separately.
First, let's find the time it takes for the shot to hit the ground. We can use the vertical motion equation:
h = h0 + (v0y * t) - (0.5 * g * t^2)
Where:
h = final vertical displacement (height of the ground) = 0
h0 = initial vertical displacement (height at which the shot was released) = 2.05 m
v0y = initial vertical velocity = v0 * sin(theta)
g = acceleration due to gravity = 9.8 m/s^2
t = time taken
Rearranging the equation, we get:
t = (v0y ± √(v0y^2 + 2 * g * h0)) / g
Since the shot is moving upwards initially, we take the positive square root in this case. Plugging in the values:
t = (16 * sin(32°) + √((16 * sin(32°))^2 + 2 * 9.8 * 2.05)) / 9.8
Now that we have the time taken for the shot to hit the ground, we can calculate the horizontal distance traveled using the equation:
d = v0x * t
Where:
d = horizontal distance traveled
v0x = initial horizontal velocity = v0 * cos(theta)
t = time taken
Plugging in the values:
d = 16 * cos(32°) * t
Now we can substitute the value of t we calculated earlier to find the horizontal distance traveled by the shot.