In the Olympic shotput event, an athlete throws the shot with an initial speed o
ID: 2134468 • Letter: I
Question
In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 m/s at a 40.0? angle from the horizontal. The shot leaves her hand at a height of 1.8 m above the ground.How far does the shot travel?
Express your answer to four significant figures and include the appropriate units.
x=______________
Part B
Repeat the calculation of part (a) for angles of 42.5 ?, 45.0 ?, and 47.5 ?.
Express your answer to four significant figures and include the appropriate units.
x(42.5?) = ___________
Part C
Express your answer to four significant figures and include the appropriate units.
x(45.0?) =______________
Part D
Express your answer to four significant figures and include the appropriate units.
x(47.5?) =____________
In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 m/s at a 40.0? angle from the horizontal. The shot leaves her hand at a height of 1.8 m above the ground.
How far does the shot travel?
Express your answer to four significant figures and include the appropriate units.
x=______________
Part B
Repeat the calculation of part (a) for angles of 42.5 ?, 45.0 ?, and 47.5 ?.
Express your answer to four significant figures and include the appropriate units.
x(42.5?) = ___________
Part C
Express your answer to four significant figures and include the appropriate units.
x(45.0?) =______________
Part D
Express your answer to four significant figures and include the appropriate units.
x(47.5?) =____________
Explanation / Answer
Known value...
X0 = 0m
T0 = 0s
Y0 = 1.8m
V0 = 12m/s
Ay = -9.8m/s^2
Angle = 40degree
V0x = V0cos(40)
V0y = V0sin(40)
Y1 = Y0 + V0y*(T1-T0) - 0.5g(T1)^2
0 = 1.8 + 12sin(40)*(T1) - 0.5(9.8)(T1)^2
0 = 1.8 + 7.71(T1) - 4.9(T1)^2
Solve the quadratic equation, you'll get 1.78s for T1
X = (V0x)(T1) = 12cos40(1.78) = 16.36m