A hiker throws a stone from live upper edge of vertical ground is 25.0 n\'s dire

Q: A hiker throws a stone from live upper edge of vertical ground is 25.0 n's dire

16 consider the motion along the vertical direction Y = vertical displacement = height Voy = initial velocity in Y-dirtection = 25 Cos40 ay = acceleration = 9.8 m/s2 t = time taken = 3.75 s Using the equation Y = Voy t + (0.5) ay t2 Y = ( 25 Cos40 ) (3.75) + (0.5) (9.8) (3.75)2 Y = 140.72 m D 141 m 17. consider the motion in horizontal direction Vox = initial velocity in X-direction = 25 Sin40 X = Vox t = (25 Sin40 ) (3.75) = 61.8 m B 61.8 m/s

ID: 1507232 • Letter: A

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A hiker throws a stone from live upper edge of vertical ground is 25.0 n's directed at 40.0 degree with the face of the cliff, as shown in the 1 3.75 s after being and feels no appreciable air resistance as it falls. What is the height of the cliff? 68 m 141 m none of these 129 m How fast is the stone moving just before it hits the ground? 0 m's 45.1 m s 61.8 ms none of these 58.2 nv The figure shows a graph of the acceleration of a 125-g object as a function of the net force acting on it. What are the accelerations at points A and B in mV? 16.4.0 50.6.5 0.5.2.0 0.25.0.06 none of these FIN) Two objects, each of weight W. hang vertically by spring scales as shown in the figure. The pulleys and the strings attached to the objects have negligible weight, and there is no appreciable friction in the pulleys. The reading in each scale is Zero D.2H' less than W. none of these Who has a greater weight to mass ratio: a person weighing 800 N or a person weighing 1.000 N? neither; their ratios are the same the person weighing 800 N the person weighing 1.000 N none of these You swing a bat and hit a heavy box with a force of 1500 N. The force the box exerts on the bat is exactly 1500 N only if the box does not move. exactly 1500 N whether or not the box moves greater than 1500 N if the box moves. greater than 1500 N if the bat bounces back. less than 1500 N if the box moves.

Explanation / Answer

consider the motion along the vertical direction

Y = vertical displacement = height

Voy = initial velocity in Y-dirtection = 25 Cos40

ay = acceleration = 9.8 m/s2

t = time taken = 3.75 s

Using the equation

Y = Voy t + (0.5) ay t2

Y = ( 25 Cos40 ) (3.75) + (0.5) (9.8) (3.75)2

Y = 140.72 m

D 141 m

17.

consider the motion in horizontal direction

Vox = initial velocity in X-direction = 25 Sin40

X = Vox t = (25 Sin40 ) (3.75) = 61.8 m

B 61.8 m/s

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A hiker starts at his camp and moves the following distances while exploring his

A hiker throws a stone from the upper edge of a vertical cliff (see the figure b

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A hiker throws a stone from live upper edge of vertical ground is 25.0 n\'s dire

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