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a ball is projected with a velocity of 40√2 m/s at an angle of 45 find the position and velocity of the ball after 2 sec

a ball is projected with a velocity of 40√2 m/s at an angle of 45 find the position and velocity of the  ball after 2 sec
 

Grade:10

2 Answers

Prabhjot Singh
19 Points
7 years ago
velocity in both x and y directions=40 m/s. For X coordinate, X=40t i.e X=40x2=80m
For Y coordinate, Y=Uyt+1/2gt^2 i.e. Y=40x2-1/2x10x4=60m
Position of ball after 2 sec= (80,60)
Velocity in X direction remains same. For Y direction, Vy=Uy+gt i.e. Vy=40-10x2=20m/s
i.e. velocity of ball =(40^2+20^2)^1/2=44.72 m/s and direction=tan-1(20/40)=tan-1(1/2)=26.56 degrees
the
10 Points
7 years ago
please explain further in an easier way 

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