Physics
posted by Mrs. Jones on .
A shortputter 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=vₒg•t1.
At the top point
0= vₒg•t1,
t1= vₒ/g=16/9.8=1.63.
h=vₒ•t1g•t1²/2 =16•1.639.8•(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.