Question

In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 44.0 ∘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?

Express your answer with the appropriate units.

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial speed the shot put is thrown at = V = 13 m/s

Angle the shot put is thrown at = = 44^o

Initial horizontal velocity of the shot put = V_x = VCos = (13)Cos(44) = 9.35 m/s

Initial vertical velocity of the shot put = V_y = VSin = (13)Sin(44) = 9.03 m/s

Initial height the shot put leaves the athletes hand = H = 1.8 m

Time taken by the shot put to reach the ground = T

When the shot put reaches the ground the vertical displacement of the shot put is downwards therefore it is negative.

T = 2.022 sec or -0.181 sec

Time cannot be negative.

T = 2.022 sec

Horizontal distance covered by the shot put = R

There is no horizontal force acting on the shot put therefore the horizontal velocity of the shot put remains constant.

R = 18.9 m

Distance the shot put travels = 18.9 m