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Phy-XI-4-14 Numerical on Projetiles
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Numerical of projectile motion and circular motion are explained in this lesson

State topper in school exams, selected in IIT Kharagpur, M.Sc., 30 yrs in teaching.

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1. PHYSICS for NEET and JEE Pradeep Kshetrapal Class XI Portion

2. Chapter 4- Motion in a Plane Numerical on 2-D motion Phy-XI-4-13

3. Q1. the ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the celiling of the hall? Solution: rw2 25 m Let 0 be the angle with which the ball is thrown in the hall

4. We need to find the maximum range after throwing the ball such that the ball does not hit the ceiling. We know that the maximum range we get from angle 6-45 . But, it is not necessary that at 45 it will not hit the ceiling. So, we check heightmax-42 (sin 0)2 2g

5. 1 12 403x ( = 40m. 2 x 10 Therefore, should be less than 45 . To find we apply 402-(sin )2 20 25 5 sin Or cos 6- V16 V 16

6. 2 u Range--x 2x sinex cos 5 11 40x40x2x 161 5 10 160 x 2 V55 16 = 20V 55-20 x 7.4 148m

7. Q2. A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground the cricketer throw the same ball. Solutions: Hmax maximum range =100m

8. 12 (sin )2 2g We know that H- 2 max We know that to attain maximum height sin should be 1. thus, for maximum height the ball should be thrown vertically upward straight making 90 with the horizontal 2 max 2g

9. Given, maximum range-100-_. u2 100_EO =-=-=50. max-2g- 2 Thus, 50m is the maximum height the ball will attain. *11

10. Q3. A stone tied to the end of a string 80cm 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? Solution: n there has to Centrisealaceration s constant 80 cm constant To move into a circular motion there has to has a centripetal acceleration acting on the stone which acts towards centre Centripetal acceleratio