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**Subjective Questions (Level 1)Projectile Motion from Ground to GroundQues 1: A particle is projected from ground with initial velocity u = 20âˆš2 m/s at Î¸ = 45Â°.Find:**(a) R, Hand T,

(b) velocity of particle after 1 s

(c) velocity of particle at the time of collision with the ground (x-axis).

Ans: (a) 80 m, 20 m, 4 s

(c) Time of flight

(c) âˆ´ Velocity of particle at the time of collision with ground.

or 5T

Leaving - ive sign which is not positive.

R = 20 x 5.46 = 109.2 m

â‡’

or 5T^{2} + 20T - 40 = 0

or T^{2} + 4T - 8 = 0

âˆ´ R = 20 x 1.46

= 29.2 m**Ques 3: A particle is projected from ground at angle 45Â° with initial velocity 20âˆš2 m/s. ****Find: ****(a) change in velocity, ****(b) magnitude of average velocity in a time interval from t = 0 to t = 3 s.****Ans: **(a) 30 ms^{-1} (vertically downwards)

(b) 20.62 ms^{-1}**Sol: **(a) Change in velocity (v_{c}) = Change in vertical velocity

(as horizontal velocity does not change).

= (u sin Î¸ - gt) - (u sin Î¸)

= - gt = - (10 x 3) m/s

= - 30 m/s

= 30 m/s (downward)

âˆ´ Displacement at time t (= 3 s)**Ques 4: The coach throws a baseball to a player with an initial speed of 20 m/s at an angle of 45Â° with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? (g = 10 m/s ^{2})**

â‡’

âˆ´

P = Horizontal component of velocity at O

âˆ´ v cos Î¸ = 60 cos 60Â°

â‡’ v cos 45Â° = 60 cos 60Â°

â‡’

For point P:

v sin 45Â° = 60 sin 60Â° + (-10) t

or

For point Q:

- v sin 45Â° = 60 sin 60Â° + (-10) t

âˆ´ t

= 8.20 s

Differentiating above equation w.r.t. time t

âˆ´

â‡’

â€¦(ii)

Differentiating Eq. (i) w.r.t. time t

Acceleration of particle

âˆ´

Eq. (iii)

i.e.,

Substituting above value of

**Give the approximate answers. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel?**

(c) the magnitude and

(d) the direction of the ballâ€™s velocity just before it hits the ground?

(b) 23 m

(c) 16.6 ms

(d) tan

(2) below horizontal

âˆ´ Initial vertical velocity at

Final vertical velocity at B (highest point)

= 0 m/s

Using v

0

â‡’

= 1.86 m

âˆ´ Maximum height attained by ball

= 9.1 + 1.86

= 10.96 m

(b) Let magnitude of vertical velocity at O (point of projection) = u

Using v

â‡’

= 14.8

Angle of projection

(c) Magnitude of velocity just before the ball hits ground

i.e., both will always remain in the same horizontal line as shown in figure.

At time t:

Vertical velocity of A

= Vertical velocity of B

= 0 + (+ g) t

= gt

At A:

or â€¦(i)

At B:

or â€¦(ii)

Multiplying Eq. (i) by Eq. (ii),

âˆ´

Distance between A and B

or

or tan Î¸ + tan Ï† = tan Î±

(Students to remember this formula)

â‡’

= 9/12

â‡’

If particle takes t time to reach point A

(i.e., vertical displacement of + h)

or gt

The above equation is quadratic in t. Two values of t will satisfy Eq. (i). One having lower value will be time (= t

âˆ´ Time to reach point Q from point P

Distance between P and Q:

2h = (u cos Î±) (t

âˆ´ 4 h

or 1 = cos

or 1 = cos

or 1 = cos

or 16 cos

â‡’ (4 cos

â‡’

(Proved)

At time t vertical velocity = v sin Î²

âˆ´ v sin Î² = u sin Î± = (- g) t â€¦(i)

Now, as horizontal acceleration will be zero.

v cos Î² = u cos Î±

Thus, Eq. (i) becomes

or u sin Î± cos Î² - u cos Î± sin Î² = gt cos Î²

or **Ques 12: A projectile aimed at a mark, which is in the horizontal plane through the point of projection, falls a cm short of it when the elevation is Î± and goes b cm far when the elevation is Î². Show that, if the speed of projection is same in all the cases the proper elevation is:****Sol: **

i.e.,

i.e.,

Adding Eqs. (i) and (ii), we have

or

â‡’ (Proved.)**Ques 13: Two particles are simultaneously thrown in horizontal direction from two points on a riverbank, which are at certain height above the water surface. The initial velocities of the particles are v _{1}=5 m/s and v_{2} = 7.5 m/s respectively. Both particles fall into the water at the same time. First particle enters the water at a point s = 10m from the bank. **

i.e.,

= 19.6 m

âˆ´

Horizontal velocity of balloon (+ bag)

Bag is released at point A.

Let t be time, the bag takes from A to reach ground.

= 3.37 s

Vertical velocity of bag when it strikes ground

âˆ´ Velocity of bag with which it strikes ground

= 37.44 m/s

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