A. h = 0.5g*t^2 = 0.90 m.
4.9t^2 = 0.90
t^2 = 0.184
t = 0.429 s. = Fall time.
B. Dx = Xo*t = 1.6 m.
Xo * 0.429 = 1.6
Xo = 3.73 m/s. = Hor. component.
C. Y = Yo+g*t = 0 + 9.8*0.429 = 4.20 m/s
= Ver. component.
V^2 = Xo^2 + Y^2 = 3.73^2 + 4.2^2=31.55
V = 5.62 m/s.
D. tan A = Y/Xo = 4.2/3.73 = 1.12601
A = 48.4o. = Direction.
tennis ball rolls off the edge of a tabletop 0.900m above the floor and strikes the floor at a point 1.60m horizontally from the edge of the table.
A: Find the time of flight of the ball
B: Find the magnitude of the initial velocity of the ball.
C: Find the magnitude of the velocity of the ball just before it strikes the floor.
D: Find the direction of the velocity of the ball just before it strikes the floor.
2 answers
Distance covered along the pathway,
s = sqrt(0.9^2 + 1.6^2) = 1.836 m
Now, s = ut + ½*gt^2
=> 1.836 = 0.5 * 9.8 * t^2 [since u=0]
=> t = 0.612 sec (Fall time)
s = sqrt(0.9^2 + 1.6^2) = 1.836 m
Now, s = ut + ½*gt^2
=> 1.836 = 0.5 * 9.8 * t^2 [since u=0]
=> t = 0.612 sec (Fall time)
5 answers