All posts related to kinematics on this blog can be accessed either by clicking on the label ‘kinematics’ below this post or by performing a search using the search box at the top of this page. Today I give you a few multiple choice practice questions with solution:

(1) A stone projected vertically up is found to be at a height *h* at times 1 s and 3 s. Neglect air resistance and assume that *g = *10 ms^{–2}. The maximum height reached by the stone is

(a) 60 m

(b) 50 m

(c) 44 m

(d) 22 m

(e) 20 m

Evidently the stone reaches the same height *h *first during its upward trip and next during its downward trip. But you need not bother about this because everything is contained in the equation of motion, *x *– *x*_{0 }= *v*_{0}*t* + ½ *at*^{2} (or, *s = ut* + ½ *at*^{2})* *

Considering the two cases given in the question, since *a = *–* g* (taking upwrd as positive and therefore downward negative),* *we have

*h* = *v*_{0}×1 – ½ ×10×1^{2} = *v*_{0}×3 – ½ ×10×3^{2} from which

2*v*_{0 }= 40 so that *v*_{0} = 20 ms^{–1}

The naximum height *h*_{max }reached is given by by 0 = *v*_{0}^{2} – 2*gh*_{max}

[This is the equation, *v*^{2 }*= v*_{0}^{2} + 2*a*(*x *– *x*_{0}) which can also be written as *v*^{2 }*= u*^{2} + 2*as*]* ^{}*

Therefore, *h*_{max} = *v*_{0}^{2}/2*g* = 20^{2}/(2×10) = 20 m

(2) A food packet is released from a helicopter ascending vertically up with uniform velocity of 2 ms^{–1}. Neglect air resistance and assume that *g = *10 ms^{–2}. The velocity of the packet at the end of 0.2 s will be

(a) 5 ms^{–1}

(b) 4 ms^{–1}

(c) 2 ms^{–1}

(d) 1 ms^{–1}

(e) zero

The initial velocity *v*_{0}* *of the packet is 2 ms^{–1} and is directed *upwards*. The gravitational acceleration is directed *downwards*. The velocity *v *of the packet at time *t* is given by

*v =** v*_{0} – *gt*.

[Note that we have taken the upward vector as positive and hence the downward vector as negative].

Substituting for *v*_{0}, *g *and *t*, *v* = 2 – (10×0.2) = 0.

[After getting released, the packet moves upwards for 0.2 s with uniformly *decreasing* speed, momentarily comes to rest and then moves downwards with uniformly *increasing* speed].

(3) An astronaut on a planet having no atmosphere throws a stone of mass *m *vertically upwards with velocity 10 ms^{–1}. It reaches maximum height of 10 m. If another stone of mass 2*m* is thrown upwards with velocity 20 ms^{–1}, the maximum height reached will be

(a) 80 m

(b) 40 m

(c) 30 m

(d) 20 m

(e) 10 m

The maximum height (*h*_{max}) reached is given by

*h*_{max} = *v*_{0}^{2}/2*g*

[Remember *v*^{2 }*= v*_{0}^{2} + 2*a*(*x *– *x*_{0}) which can be written as *v*^{2 }*= v*_{0}^{2} – 2*gh*_{max } for projection vertically apwards]* ^{}*

Substituting for *h*_{max} (= 10 m) and *v*_{0} (= 10 ms^{–1}), the acceleration due to gravity (*g*) on the planet is obtained as 5 ms^{–2}.

For a given velocity of projection, the height reached is *independent* of the mass of the stone.

When the velocity of projection is changed to 20 ms^{–1}, the maximum height reached is 20^{2}/(2×5) = 40 m.

[You may note that the maximum height is directly proportional to the

The following question is specifically for AP Physics C aspirants. But AP Physics B aspirants too can ‘enjoy’ it.

(4) A ball thrown vertically up from the ground moves very close to an open window of a room in the first floor of a building. A student in the room sees the ball through the window, when the ball moves up as well as when it move down. The height of the window is 2 m and the ball is *in view* for a total time of 0.2 s. Take *g* = 10 ms^{–2}.^{ }How high above the top of the window does the ball go (approximately)?

(a) 19 m

(b) 20 m

(c) 21 m

(d) 24 m

(e) 25 m

Since the total time for which the ball is in view is 0.2 s, the time taken by the ball to traverse the height of the window is 0.1 s. The average upward velocity (*v*) of the ball while moving past the window is 2/0.1 = 20 ms^{–1}. We may take this to be very nearly equal to the velocity of the ball at the *centre* of the window.

Therefore, the maximum height reached by the ball (from the centre of the window) is given by

*h*_{max} = *v*^{2}/2*g* = 20^{2}/(2×10) = 20 m.

[We have used the equation, *v*^{2}* =u*^{2} + 2*as* where *a* =* *– *g* and *s* = *h*_{max}].

Therefore, the ball goes to a height of nearly 19 m above the top of the window [Option (a)].

[Since the motion of the ball is accelerated, the average velocity is not exactly equal to the velocity at the centre of the window. The answer we obtain is an *approximate* one.

If you wnt to calculate the *exact* height, you will proceed as follows:

If *t* is the time taken by the ball to fall from the topmost point of the trajectory to the top of the window, the maximum height* h*_{max} (measured from the top of the window) is given by

*h*_{max} = 0 + ½ *gt*^{2} = 5* t*^{2} ………(i)

Also, (*h*_{max} + 2) = 0 + ½ *g *(*t *+0.1)^{2} = 5*t*^{2} + 5×0.01 + 5×2×0.1*t*

Or, *h*_{max} + 2 = 5* t*^{2} + 0.05 + *t* ……….(ii)

Subtracting Eq (i) from Eq (ii) we obtain *t = *1.95 s.

Therefore, *h*_{max} = 5* t*^{2} = 5×(1.95)^{2} = 19.0125 m.

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