A child drops a ball from a bridge. The ball strikes the water under the bridge 2.0 seconds later.
What is the velocity of the ball when it strikes the water?
Solution
V=U+AT=0+32*2=64 FT/SEC...OR IN MKS UNITS....
V=0+9.8*2=19.6 M/SEC
SEE BELOW FOR EXPLANATION
A boy throws a rock straight up into the air. It reaches the highest point of its flight after 2.5
seconds. How fast was the rock giong when it left the boys hand?
1 solutions
Answer 17643 by venugopalramana(1171) About Me on 2006-03-21 11:25:11 (Show Source):
THE FORMULA TO BE USED IS
V=U+AT...WHERE
V=FINAL VELOCITY....FT/SEC.
U=INITIAL VELOCITY....FT/SEC.
A = ACCELERATION ..HERE IT IS DUE TO GRAVITY WHICH IS ACTING AGAINST
THE MOTION ..ITS VALUE IS -32 FT/SEC.SEC.
T=TIME OF TRAVEL...SEC
WE HAVE U=?
V=0....AS IT REACHED THE HIGHEST POINT AND WILL STOP AND FALL BACK
FROM THERE.
T=2.5 SEC.
0=U-32*2.5
U=32*2.5=80 FT/ SEC.
IF YOU ARE IN MKS UNITS THEN A=-9.8 M/SEC.SEC
U=9.8*2.5=24.5 M/SEC.
A child drops a ball from a bridge. The ball strikes the water under.pdf
1. A child drops a ball from a bridge. The ball strikes the water under the bridge 2.0 seconds later.
What is the velocity of the ball when it strikes the water?
Solution
V=U+AT=0+32*2=64 FT/SEC...OR IN MKS UNITS....
V=0+9.8*2=19.6 M/SEC
SEE BELOW FOR EXPLANATION
A boy throws a rock straight up into the air. It reaches the highest point of its flight after 2.5
seconds. How fast was the rock giong when it left the boys hand?
1 solutions
Answer 17643 by venugopalramana(1171) About Me on 2006-03-21 11:25:11 (Show Source):
THE FORMULA TO BE USED IS
V=U+AT...WHERE
V=FINAL VELOCITY....FT/SEC.
U=INITIAL VELOCITY....FT/SEC.
A = ACCELERATION ..HERE IT IS DUE TO GRAVITY WHICH IS ACTING AGAINST
THE MOTION ..ITS VALUE IS -32 FT/SEC.SEC.
T=TIME OF TRAVEL...SEC
WE HAVE U=?
V=0....AS IT REACHED THE HIGHEST POINT AND WILL STOP AND FALL BACK
FROM THERE.
T=2.5 SEC.
0=U-32*2.5
U=32*2.5=80 FT/ SEC.
IF YOU ARE IN MKS UNITS THEN A=-9.8 M/SEC.SEC
U=9.8*2.5=24.5 M/SEC
