Question

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0 m/s at a 40.0 ∘ angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground.

Answer 1

Answer:

Explanation:

given,

initial speed of the shot = 12.0 m/s

angle = 40°

height at which shot leaves her hand = 1.80 m

v_x = 12 cos 40° = 9.19 m/s

v_y = 12 sin 40° = 7.71 m/s

time to reach maximum height =

     = $\dfrac{v_y}{g}$

     = $\dfrac{7.71}{9.8}$

     = 0.787 s

$h = v_y t + \dfrac{1}{2}gt^2$

h = 7.71 × 0.787 - 0.5 × 9.81 × 0.787²

h = 3.03 m

the maximum height attain = 3.03 + 1.8 = 4.83 m

now free fall from the maximum height

$h =\dfrac{1}{2}gt^2$

$4.83 = \dfrac{1}{2}\times 9.8 \times t^2$

t = 0.9928 s

total time = 0.9928 + 0.787 = 1.7798 s

range =

d = vₓ t

d = 16.36 m