A lead ball is dropped in a lake from a diving board 5.20 m ahove the water. It hits the water

A lead ball is dropped in a lake from a diving board 5.20 m...

Key Concepts

Multi-Phase Motion Analysis

Many motion problems involve different phases, such as a period of accelerated motion followed by constant velocity motion. Multi-phase motion analysis involves breaking down the overall movement into distinct parts, analyzing each segment separately with the appropriate kinematic principles, and then combining the results to understand the complete motion.

Constant Velocity Motion

Constant velocity motion refers to movement at a uniform speed in a fixed direction. In problems where an object moves with a constant velocity after an initial phase of acceleration, this concept allows for the straightforward calculation of displacement by multiplying the constant speed by the time duration of this phase.

Projectile Motion with Initial Velocity

This concept involves the motion of an object that is projected with an initial velocity and then moves under the influence of gravity. In one-dimensional vertical motion, the object’s path can be analyzed using kinematic equations to determine parameters like the initial velocity required to cover a known displacement in a given time, taking into account the constant acceleration due to gravity.

Kinematic Equations

Kinematic equations are formulas that relate displacement, initial velocity, final velocity, acceleration, and time for objects moving with constant acceleration. They enable solving for unknown quantities given others, making them essential for analyzing motion in free-fall situations and when additional velocities are involved during launching or sinking.

Uniformly Accelerated Motion

This concept describes the motion of an object when its acceleration remains constant throughout its journey. In many physics problems involving gravity, the acceleration due to gravity is treated as a constant, allowing the use of simplified equations to determine displacement, velocity, and time. This concept is crucial in understanding how objects accelerate downward when dropped or thrown.

Average Velocity

Average velocity is defined as the total displacement divided by the total time taken for the journey. It provides a single value that represents the overall rate at which an object’s position changes, regardless of any variations in speed during different segments of the motion. This is useful for comparing overall motion with instantaneous velocities at certain points in time.

Transcript

00:01 In this problem of motion along a straight line we have given that a lead ball is dropped in a lake from a diving board this is 5 .2 meters above the water.

00:11 Say this is the height of diving board above the water that is 5 .2 meters.

00:16 It hits the water with a certain velocity and then sinks to the bottom with the same constant velocity.

00:24 It reaches the bottom.

00:25 So bottom it reaches this is 4 .80 seconds after it is dropped.

00:33 We have to find how deep is the lake.

00:35 So let's draw a condition say this is the diving boat and this is the, we can say this is the water surface and this is the earth level or we can say the ground.

00:49 So this is ground.

00:52 When we drop, say this is lead ball.

00:55 So this would be in constant acceleration until it reaches the water and then from here it dropped with a constant velocity say this is point o and this is point a and this is b.

01:10 Here we have v is equal to constant and here we would have acceleration is equal to gravitational constant this is g which is equal to in negative direction so we can say this is negative 9 .8 meters per second square and this is the initial point now we have to find the depth say this is d and let's suppose the time taken from o to b is equal to t1 and here this value would be t2 now we have to find the velocity at point a so velocity at point a would be say v is equal to under root of 2g h and this h is equal to 5 .2 meters so this is given that this is 5 .20 meters putting the value so we get v is equal to 2 multiplied with 9 .8 multiplied with 5 .2 when we solve it, this is.

02:18 So this is equal to approximately this is 10 meters per second.

02:23 Now we have to find the deep, how deep is the lake? so we say that time taken, say t1 plus t2.

02:32 So we have t1 plus t2 is equal to 4 .80 seconds.

02:37 So this is 4 .80 second.

02:39 So better is to find the value of t1.

02:41 So t1 is calculated as, say t1 is calculated as, under root of 2h divided with the constant this is gravitational constant say g so this value would be under root of 2 multiplied with 5 .20 and divided with this is g which is 9 .8 meters per second square and this would be in seconds and t2 t2 would be calculated as say here t2 would be calculated as 4 .80 minus t1 so 4 .80 minus we have to find the value of t1 so is equal to so this is 1 .03 seconds now 4 .80 minus 1 .03 is equal to 3 .77 so this is equal to 1 .03 so this is equal to 3 .77 seconds now it goes with the constant velocity and here velocity is equal to 10 meters per second so this is 10 meters per second and actually this is 10 .01.

03:48 So when we solve it this is d is equals to v t so velocity multiplied with time so well city is 10 .01 or we can say this is roughly 10 multiplied with t which is equal to 3 .77 so this is 3 .77 now when we solve it this is equal to 37 .7 or we can say this is approximately equals to 38 meters so we have the depth of water is equal to or we can say the width of lake is equal to 38 meters.

04:23 This is a solution for part a and now for part b.

04:27 Part b says we have to find the magnitude and direction of the average velocity of the ball for the entire fall.

04:37 So average velocity is given as, say, v average is given as, say, height plus d, say h plus d divided with the time taken total...

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