Question

12 A stone of mass 0.20 kg is thrown with speed 22 m s-1 from the edge of a cliff that is 32 m above the sea. The initial velocity of the stone makes an angle of 35° with the horizontal. Air resistance is neglected. a i Determine the horizontal and vertical components of the initial velocity. [2] ii Sketch graphs showing the variation with time of the horizontal and vertical components of velocity. [2] b i Calculate the maximum height above the cliff reached by the stone. [3] ii State the net force on the stone at the highest point in its path. [1] c i Using conservation of energy, determine the speed of the stone as it hits the sea. [2] ii Hence or otherwise, determine the time it took the stone to reach the surface of the sea. [2] The graph shows the path followed by this stone, until just before hitting the sea, in the absence of air resistance. d i On a copy of the axes above, draw the path of the stone in the presence of an air resistance force opposite to the velocity and proportional to the speed. [3] ii State and explain one difference between your graph and the graph above. [2]

          12 A stone of mass 0.20 kg is thrown with speed 22 m s-1 from the edge of a cliff that is 32 m above the sea. The initial velocity of the stone makes an angle of 35° with the horizontal. Air resistance is neglected.
a i Determine the horizontal and vertical components of the initial velocity. [2]
ii Sketch graphs showing the variation with time of the horizontal and vertical components of velocity. [2]
b i Calculate the maximum height above the cliff reached by the stone. [3]
ii State the net force on the stone at the highest point in its path. [1]
c i Using conservation of energy, determine the speed of the stone as it hits the sea. [2]
ii Hence or otherwise, determine the time it took the stone to reach the surface of the sea. [2]
The graph shows the path followed by this stone, until just before hitting the sea, in the absence of air resistance.
d i On a copy of the axes above, draw the path of the stone in the presence of an air resistance force opposite to the velocity and proportional to the speed. [3]
ii State and explain one difference between your graph and the graph above. [2]

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12 A stone of mass 0.20 kg is thrown with speed 22 m s-1 from the edge of a cliff that is 32 m above the sea. The initial velocity of the stone makes an angle of 35° with the horizontal. Air resistance is neglected.
a i Determine the horizontal and vertical components of the initial velocity. [2]
ii Sketch graphs showing the variation with time of the horizontal and vertical components of velocity. [2]
b i Calculate the maximum height above the cliff reached by the stone. [3]
ii State the net force on the stone at the highest point in its path. [1]
c i Using conservation of energy, determine the speed of the stone as it hits the sea. [2]
ii Hence or otherwise, determine the time it took the stone to reach the surface of the sea. [2]
The graph shows the path followed by this stone, until just before hitting the sea, in the absence of air resistance.
d i On a copy of the axes above, draw the path of the stone in the presence of an air resistance force opposite to the velocity and proportional to the speed. [3]
ii State and explain one difference between your graph and the graph above. [2]

Added by Melanie S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Rahul Kumar

11/18/2022

Step 1

- The initial velocity of the stone is \(22 \, \text{ms}^{-1}\) at an angle of \(35^\circ\) with the horizontal. - To find the horizontal component (\(v_{x0}\)), use the formula: \(v_{x0} = v_0 \cos(\theta)\), where \(v_0 = 22 \, \text{ms}^{-1}\) and \(\theta = Show more…

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A stone of mass 0.20 kg is thrown with speed 22 m s-1 from the edge of a cliff that is 32 m above the sea. The initial velocity of the stone makes an angle of 35° with the horizontal. Air resistance is neglected.

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00:01 In this question, a 0 .2kg mass is thrown with a speed of 22 meter per second and the angle made with the horizontal is 35 degree.

00:14 So let's calculate the vertical and horizontal component of initial velocity.

00:20 So horizontal component of initial velocity will be v cost 35, that is equals to 22 into cost 35.

00:29 And the vertical component will be v.

00:35 Sorry, v.

00:36 Sine 35, so that is equals to 22 into sine 35.

00:44 So calculating this we will get v.

00:49 Cost 35 equals to 18 meter per second.

00:53 So ux will be equals to 18 meter per second and the vertical component will be equals to 12 point or u.

01:09 Will be equal to 12 .6 meter per second that is in horizontal direction the velocity will be 18 meter per second and in vertical direction it will be 12 .6 meter per second so that is the answer for option a now when we draw the sketch for the variation of time with the with velocity let's say this this is ux, that is the horizontal component of velocity and on x on x axis, this is time.

01:50 So the graph will be something like this, that is with passing of time, the horizontal component of velocity doesn't change, it remains same.

02:03 Whereas in the case of u y and time when we draw the graph it will be something like this way.

02:25 The graph will be something like this way.

02:29 Basically the governing equation for uy is u0y that is the initial velocity, vertical component velocity minus gt.

02:39 So when the time passes this value go on increasing.

02:45 So basically what happens that let's say if this is the graph, this is the path of the projectile.

02:54 So initially the velocity goes on with a very high velocity and it goes on decreasing decreasing and here when you can see that it becomes zero at this point it goes on also in negative direction so this will be the graph for the vertical component of velocity and here the governing equation will be 12 .6 minus 9 .8 t and here ux will be remains constant and that is 18 meters per second.

03:40 So this is the answer for part a.

03:43 Now coming to the another part that is part b we have to calculate the maximum height.

03:50 Now at maximum height the velocity becomes zero...

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