Question

In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 47.0° angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. How far does the shot travel?

Answer 1

Answer:

18.7 m

Explanation:

$v_{o}$ = initial speed of the shot = 13 m/s

θ = angle of launch from the horizontal = 47 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 1.80 m

t = time of travel

Using the kinematics equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 1.80 = (9.5) t + (0.5) (- 9.8) t²

t = 2.11 s

Consider the motion along the horizontal direction

x = horizontal displacement of the shot

$v_{ox}$ = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.11 s

Using the kinematics equation

$x=v_{ox} t+(0.5)a_{x} t^{2}$

x = (8.87) (2.11) + (0.5) (0) (2.11)²

x = 18.7 m