A shot-putter throws the shot ( mass = 7.3 kg) with an initial speed of 14.0 m/s at a 33.0 ∘ angle to the horizontal.

Question

Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 1.80 m above the ground.

Δx =

Answer 1

ΔΔΔSolution:

In projectile motion there are two components to a force shot at an angle. The vertical component and the horizontal component . The horizontal component X = V_ocos ∝ t where ∝ = 33º the angle with the horizontal and V_o is the initial velocity given in this problem as 14.0 m/s

and t is the time it takes to travel in seconds . the challenge here is to find t

t can be found from the vertical component y = -1/2 gt² + V₀sin∝ t + y₀ recognizing that y_o is the initial height and it is equal to 1.8m. the final height of the shot-put is y =0 when it is on the ground, the vertical force component is V_osin ∝ t, and g is the gravitational acceleration of 9.8 m/s ². Then putting all that together in the vertical equation of motion you get:

0 = -1/2 ( 9.8 m/s²) t² + ( 14 m/s)( sin(33º) t + 1.8 m , and solving for t:

0 = -4.9t² + 7.625 t + 1.8 , rearranging 4.9t² -7.625 t -1.8 = 0

This is a quadratic equation with two roots with a = 4.9 , b= -7.625 , and c = -1.8

solving Δ = b² - 4ac = ( -7.625)² -4(4.9)(-1.8) = 93.4 , t = -b ± √Δ / 2a = [-(-7.625) ± √ 93.4] / 2(4.9)

t = (7.625 ± 9.66) / 9.8 and is either t = (7.625 + 9.66) / 9.8 or t = (7.625 - 9.66) /9.8

t = 1.76 s or t = -0.21 s realizing the time cannot be negative value then the 2nd answer is disregarded

and t = 1.76 s

the travelled distance Δx = V_ocos ∝ t = (14 m/s)(cos(33°)(1.76s) = 20.7 m