A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent an

step 1 In this problem of motion along a straight line we have given that. A basketball player grabbing

A basketball player grabbing a rebound jumps 76.0 cm...

Key Concepts

Projectile Motion

Projectile motion covers the analysis of objects moving under the influence of gravity after an initial launch. In vertical motion scenarios, the time of ascent and descent can be computed using kinematic equations, and the symmetry of upward and downward flight under constant gravity is often exploited.

Segment Analysis in Vertical Motion

Segment analysis involves breaking the overall motion into distinct parts or regions – such as the top and bottom segments of a jump. Each segment can be separately analyzed to determine parameters like time spent, using the kinematic equations, with careful attention to the changing velocities within each part of the motion.

Constant Acceleration Kinematics

This concept involves the use of equations that describe the motion of an object under a constant acceleration, such as gravity. It allows one to relate displacement, initial velocity, time, and acceleration, which is particularly useful when analyzing different segments of a projectile’s trajectory.

Gravitational Acceleration

Gravitational acceleration is the constant rate at which objects accelerate towards the Earth in free fall. This value is central to problems involving vertical motion, such as jumps or free-fall, where it defines how quickly the velocity changes over time.

Transcript

00:01 In this problem of motion along a straight line we have given that.

00:04 A basketball player grabbing a rebound jumps 76 cm vertically.

00:10 So this is 76 .0 centimeters vertically.

00:15 How much total time, ascent and descent does the player spend? so first in the top 15 cm of this jump and in the bottom 15 cm of this jump.

00:30 So first we have to find the we can say this is y vertical distance is 76 .0 centimeter or we can say this is 0 .76 meters.

00:42 Now we have to find the initial velocity so v0 is equal to under root of 2g.

00:46 Y so this is equals to under root of 2 and g is equal to 9 .8 meter per second square and y is equal to 0 .76 and this value would be in meters per second.

00:58 So, v0 comes to be 3 .86 meters per second.

01:04 That means this is the initial velocity of the basketball player leaving the ground.

01:09 So condition would be like this, say this is the x and y x.

01:15 So this would be the condition.

01:20 This is the top 76 cm or we can say the 76 cm actually, 0 .76 meter.

01:30 Now this is the time of ascend and descent in top.

01:35 15 cm and here we have top and this is bottom so this is bottom 15 centimeter now we have to find say this is we can say this would be t1 and this value would be again t1 and this is t2 so this value would be again t2 and now we can write it this is 0 .61 meters so this is 0 .61 meters and this point is here 15 centimeters or we can say 0 .15 meters so this is 15 centimeter and this is 76 centimeter now we have to find the velocity at this point at 0 .61 meters so we can say velocity at 0 .61 meters this would be equals to we can name it as this is v1 so we have v0 square minus v1 is square equals 2g and this is 0 .61.

02:44 So this is 2g multiplied with 0 .61.

02:48 Now putting all the values, say v0 is equal to 3 .86 meter per second, putting all the values so we have v1 is equal to under root of v0 square minus 2g and multiplied with 0 .61 and this value would be in meters per second...

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