The coach throws a baseball to a player with an initial speed of 20 m//s at an angle of 45^@ with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? (g = 10 m//s^2).
Answers
Given :
initial velocity of ball = 20 m/s
angle of projection = 45°
Distance between player & coach is 50 metres
To find : In which direction & by how much should player move in order to catch the ball .
Solution :
•When a body is projected with some initial velocity & it moves under the effect of gravity , this motion is known as projectile motion.
•Distance covered by projectile horizontally under influence of gravity is Known as Range
•Also Range =( u•² sin2Q)/g
Where , u• = initial velocity of ball
Q = Angle of projection of ball
g = Acceleration due to gravity
•In case of ball
Range = { (20)²sin(2)(45) }/10
Range = [ 400×sin(90°) ]/10
Range = 40 m
=> distance covered by ball will be 40 metres
•So, player has to move 10 metres towards coach to catch the ball at from the same level in which it was thrown.