A track meet is held on a planet in a distant solar system. A shot-putter releas

Question

A track meet is held on a planet in a distant solar system. A shot-putter releases a shot at a point 2.0 m above ground level. A stroboscopic plot of the position of the shot is shown in the figure, where the readings are 0.58 s apart and the shot is released at time t 0. (a) What is the initial velocity of the shot in unit-vector notation? (b) What is the magnitude of the free-fall acceleration on the planet? (c) How long after it is released does the shot reach the ground? (d) If an identical throw of the shot is made on the surface of Earth, how long after it is released does it reach the ground? 10 t 0 15 30 x (m) J Units (a) Number (b) Number Units (c) Number Units (d) Number Units Click if you would like to show work for this question Open Show Work olicy I Q 2000-201 ohn Wiley & Sons Inc. All Rights Reserved. A Division of John Wiley & Sons, Inc. Version 4.22.1.2

Explanation / Answer

horizontal range ,R = ux*t = 25

So, ux*(0.58*5) = 25

So, ux = 8.62 m/s

In vertical direction : s = uy*t - 0.5*g*t^2

So, 6 = uy*0.58 - 0.5*g*0.58^2 ------ (1)

and 8 =  uy*(2*0.58) - 0.5*g*(2*0.58)^2 ------(2)

Solving both : uy = 13.8 m/s and g = 11.9 m/s2

So, initial velocity = 8.62 i + 13.8 j <-------------answer

b)

free fall acceleration, g = 11.9 m/s2 <------ solved in previous part

c)

s = ut + 0.5*at^2

So, -2 = 13.8*t - 0.5*11.9*t^2

So, t = 2.46 s <----answer

d)

On earth , g = 9.8 m/s2

So, -2 = 13.8*t - 0.5*9.8*t^2

So, t = 2.95 s <----answer

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