Physics
posted by Mrs. Jones on .
A shotputter 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?

Upwards motion
v=vₒg•t1.
At the top point
0= vₒg•t1,
t1= vₒ/g=16/9.8=1.63.
h=vₒ•t1g•t1²/=16•1.639.8•(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.