Question

Problem 3.29 A shot-putter throws the "shot" (mass = 7.3 kg ) with an initial speed of 14.5 m/s at a 37.0 ∘ angle to the horizontal.

Part A Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.15 m above the ground. Express your answer using three significant figures and include the appropriate units. d =

Answer 1

m = 7.3 kg
Vo = 14.5 m/s
Yo = 2.15 m
θ = 37.0°
g = 9.81 m/s^2 ---> Earth gravity

Calculate initial horizontal and vertical components of velocity

Vox = Vo * cosθ
Vox = (14.5m/s) * (cos37.0)
Vox = (14.5 m/s) * (0.765)
Vox = 11.09 m/s

Voy = Vo * sinθ
Voy = (14.5m/s) * (sin37.0)
Voy = (14.5m/s) * (0.643)
Voy = 9.33 m/s

Determine the time that it will take to rise to its maximum height from the shot-potter's arm

Tr = Voy / g
Tr = (9.33 m/s) / (9.81 m/s^2)
Tr = 0.951 s

Determine what the total height is

H = Yo + [Voy * Tr] - [0.5 * g * Tr^2]
H = [2.15m] + [ (9.33m/s) * (0.951 s) ] - [ 0.5 * (9.81 m/s^2) * (0.951s)^2 ]
H = [2.15m] + [ 8.87 m] - [ (4.905 m/s^2) * (0.904 s^2) ]
H = [2.15m] + [ 8.87 m] - [ 4.43 m]
H = 6.59 m

Calculate the time that it takes to fall from this height

Tf = SQRT { 2H / g }
Tf = SQRT { [2 * 6.59 m] / (9.81 m/s^2) }
Tf = SQRT { [13.18 m] / (9.81 m/s^2) }
Tf = SQRT { 1.34s^2 }
Tf = 1.15 s

Find the total travel time

Tt = Tr + Tf
Tt = (0.951s) + (1.15 s)
Tt = 1.101s

Now you can find distance, using horizontal component of velocity

R = Vox * Tt
R = (11.09 m/s) * (1.101s)
R = 12.21m