Motion under gravity By ghumare s m

Sanjivani College of Engineering Kopargaon

1. Unit-5 Rectilinear Motion of Particle Sub- Engg. Mechanics B y: Mr. Ghumare S. M. Motion Under Gravity When the particle projected vertically in the air, then its motion is under the gravitational force, this motion is called as Motion under Gravity. Here,

2. Motion Under Gravity Equations of Rectilinear Motion Equations for Motion Under Gravity modified from above equations

3. Motion under Gravity Important Points When objects thrown vertically upward in the air, it attains the maximum height at certain point where final velocity becomes zero (V=0). When objects released from the top of tower or building vertically downward, then its initial velocity at that point becomes zero. The time of travel required for upward and downward journey for the object must be same. Hence, time of flight = Total time for which object remained in air = Time of upward plus downward journey

4. Motion under Gravity Important Points Assume acceleration due to gravity ‘g’ as positive for downward journey and Negative for upward journey The velocity of the particle at height must be same in magnitude in upward and downward motion/ direction.

5. Numerical Example-1 Ex. From, top of the tower 100m high , a stone is dropped down. At the same time, another stone is thrown upward from the foot of the tower with a velocity 30 m/sec. When and Where the two stones will cross each other? Find the velocity of two stones at the time of crossing. Given, Ht. of tower 100m Stone 1 dropped from top, Hence initial velocity = 0 Stone 2 thrown upward with initial velocity = 30 m / s Let, ‘t’ be the time required for the stones during which both stones will covers some distance, let us say and . But, here

6. Example 1 Continue…. Distance travelled by 1 st Stone in downward direction is as follows,

7. Example 1 Continue…. Distance travelled by 2 st Stone in upward direction is as follows,

8. Example 1 Continue…. Total height of the tower is 100 m In time t both the stones have traveled distance h1 and h2,

9. Numerical Example-2 Ball ‘A’ is released from the rest at a height 12 m rom top, At the same time, another ball is thrown upward 1.5m from the ground. If the balls cross each other at a height of 6m, Determine the speed at which ball B was thrown upward. Given, Ht. of tower 12m Ball 1 dropped from top, at rest, initial velocity = 0 Stone 2 thrown upward 1.5m from ground, Find Let, ‘t’ be the time required for the Balls during which both stones will covers some distance, let us say and . But, here

10. Example 2 Continue…. Downward distance travelled by Ball A is 6m, Downward distance travelled by Ball B =4.5M

11. Thank You