A basketball player shoots at the basket, which is 4 meters...

Transcript

00:01 In this problem a basket ball is shot at an angle of 53 degree with the horizontal and it reaches a basket which is 4 meter away and 2 meter above the hand level and we have to find out the initial velocity of the ball to hit so here we will take the vertical direction as positive y axis in vertically upward direction and direction to the right as positive x -axis so when the ball reaches the basket the horizontal displacement is delta x equal to 4 meters because the basket is 4 meter away from the hand and the vertical displacement delta y is equal to 2 meters for the ball so first we consider the vertical motion and we will use the equation of kinematics delta y equal to v0 y t plus half a y t square delta y is the vertical displacement from initial to final position for the ball v0 y is initial vertical velocity which is v0 sine theta v0 is initial velocity and theta is the angle of launch which is 53 degree t is time taken to reach the basket a y is the vertical acceleration along y axis which is equal to acceleration due to gravity but it is in downward direction so it will be negative minus g t square so we have delta y equal to v0 sine theta t minus half g t square this is equation one for horizontal motion we will use again the equation of kinematics delta x equal to v x t plus half a x t square delta x is the horizontal displacement v0 x is initial horizontal velocity which is v0 cost theta times t plus a x is the horizontal acceleration which is zero for the ball because it only has vertical acceleration equal to acceleration due to gravity so horizontal acceleration is zero so this whole term will be zero and we get delta x equal to v0 cost theta times t now delta x is the horizontal displacement which is equal to four meters so from this we get time t equal to four meters upon v0 cost theta now we put the value of time t in equation one so equation one is delta y equal to v0 sine theta t and for t we write 4 meter upon v0 cos theta minus half g t square so we write in place of t 4 meter upon v0 cos theta square now delta y is the vertical displacement of the ball which is 2 meters and v0 cancels out and we get 4 meter times sine theta upon cos theta which is 10 theta and theta is equal to 53 degree which is the angle of launch minus half g is 9 .8 meter per second square and 4 meter square is 16 upon v0 square cos square theta the theta is 50.

04:10 Degrees so from this we get 2 equal to 41053 degree which is 5 .308 minus 216 .47 upon v0 square from this we get so v0 square will be equal to 216 .47 upon 5 .308 minus 2 and to find v0 we square root both sides and we get 8 .1 meter per second as the initial velocity...

Question

A basketball player shoots at the basket, which is 4 meters away and 2 meters above the hand level. Since the shooting angle 53°, what should be the initial velocity of the ball to hit?

Question

A basketball player shoots at the basket, which is 4 meters away and 2 meters above the hand level. Since the shooting angle 53°, what should be the initial velocity of the ball to hit?

Please give Ace some feedback