A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

**A. ** 76 m

**B. ** 60 m

**C. ** 65 m

**D. ** 50 m

**Answer : ****Option D**

**Explaination / Solution: **

Maximum horizontal distance,* R *= 100 m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e.,*θ* = 45°.

The horizontal range for a projection velocity*v*, is given by the relation:

*R* =( *u*2 Sin 2θ / g)

100 =

*=> *= 100 ….(i)

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity*v* is zero at the maximum height *H*.

Acceleration,*a = –g*

Using the third equation of motion:

*v*2 – *u*2 = -2*gH*

Maximum horizontal distance,

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e.,

The horizontal range for a projection velocity

100 =

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity

Acceleration,

Using the third equation of motion:

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A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

**A. ** 10.2 m , along the radius at every point towards the center

**B. ** 11.9 m , along the radius at every point away from the center

**C. ** 9.9 m , along the radius at every point towards the center

**D. ** 9.9 m , along the radius at every point away from the center

**Answer : ****Option C**

**Explaination / Solution: **

Length of the string,*l* = 80 cm = 0.8 m

Number of revolutions = 14

Time taken = 25 s

Frequency,

Angular frequency,

=

Centripetal acceleration,*ac* = ω2*r*

=

= 9.9 ms-2

The direction of centripetal acceleration is always directed along the string, towards the centre, at all points.

Length of the string,

Number of revolutions = 14

Time taken = 25 s

Frequency,

Angular frequency,

=

Centripetal acceleration,

=

= 9.9 ms-2

The direction of centripetal acceleration is always directed along the string, towards the centre, at all points.

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An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

**A. ** 10 g

**B. ** 6.4 g

**C. ** 5 g

**D. ** 23 g

**Answer : ****Option B**

**Explaination / Solution: **

Radius of the loop, *r* = 1 km = 1000 m

Speed of the aircraft, *v* = 900 km/h

=

Centripetal acceleration,

=

Acceleration due to gravity, g = 9.8 m/s2

= 6.4 g

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Which of the following statements not true?

**A. ** The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

**B. ** The net acceleration of a particle in uniform circular motion is always along the radius of the circle towards the centre

**C. ** The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

**D. ** The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

**Answer : ****Option D**

**Explaination / Solution: **

The net acceleration of a particle in circular motion is along the radius of the circle towards the centre Only in case of uniform circular motion.

The net acceleration of a particle in circular motion is along the radius of the circle towards the centre Only in case of uniform circular motion.

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The position of a particle is given by Find the velocity and acceleration of the particle.

**A. ** 3.0i^−4.0tj^ , -4.0j^

**B. ** 4.0i^−4.0tj^ , -3.0j^

**C. ** 2.0i^−4.0tj^, -2.0 j

**D. ** 5.0i^−4.0tj^, -3.0 j

**Answer : ****Option A**

**Explaination / Solution: **

Position vector

We know velocity is given by

So

Acceleration is given by

So,

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A particle starts from the origin at t = 0 s with a velocity of 10.0 m/s and moves in the x-y plane with a constant acceleration of ( 8.0+ 2.0) m . At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

u = uxi + uyj = 10j,

ux = 0 and uy = 10 m/s

a = axi + ayj = 8i + 2j,

ax = 8 and ay = 2 m/s2

x = uxt +axt2

=> 16 =

t = 2 sec

y = uyt + = 10x2 + = 24 m

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An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30, what is the speed of the aircraft?

**A. ** 172 m/s

**B. ** 192 m/s

**C. ** 193 m/s

**D. ** 182 m/s

**Answer : ****Option D**

**Explaination / Solution: **

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m

Angle subtended between the positions, ∠POQ = 30°

Time = 10 s

In ΔPRO:

tan 15° =

PR = OR tan 15°

=

ΔPRO is similar to ΔRQO.

∴PR = RQ

PQ = PR + RQ

= 2PR = 2 × 3400 tan 15°

= 6800 × 0.268 = 1822.4 m

∴ Speed of the aircraft =

= 182.24 m/s m/s

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A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m).

Height of the fighter plane = 1.5 km = 1500 m

Speed of the fighter plane, *v *= 720 km/h = 200 m/s

Let *θ *be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, *u* = 600 m/s

Time taken by the shell to hit the plane = *t*

Horizontal distance travelled by the shell = u*x**t*

Distance travelled by the plane = *vt*

The shell hits the plane. Hence, these two distances must be equal.

uxt = vt

u Sin θ = v

Sin θ =

θ = Sin-1(0.33) = 19.50

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (*H*) higher than the maximum height achieved by the shell.

∴ H =

=

= 16006.482 m

≈ 16 km

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A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

**A. ** 0.84 m , 58.5with the direction of velocity

**B. ** 0.82 m , 59.5with the direction of velocity

**C. ** 0.86 m , 54.5with the direction of velocity

**D. ** 0.85 m , 56.5with the direction of velocity

**Answer : ****Option C**

**Explaination / Solution: **

Speed of the cyclist, v = 27 km/h = 7.5 m/s

Radius of the circular turn, *r* = 80 m

Centripetal acceleration is given as:

ac = 0.7 ms-2

The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.

Since the angle between ac and aT is 900, the resultant acceleration *a* is given by:

a =

= = 0.86 ms-2

tan θ =

where θ is the angle of the resultant with the direction of velocity.

tan θ = = 1.4

θ = tan-1(1.4) = 54.56° with the direction of velocity

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A ball is thrown upwards with an initial velocity of 10 m . Determine the maximum height reached above the thrower’s hand. Determine the time it takes the ball to reach its maximum height.

Initial velocity u = 10m/s

As at the maximum height ball ll stop so final velocity v = 0 m/s

Only acceleration working on it is acceleration due to gravity g = -9.8m/s2

Let height = h

So we know

=>

=> 5.10 m

Also let time taken to reach maximum height = t

Then

We know

v = u +at

=> 0 = 10 +(-9.8)t

=> t =

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