1) d=v(avg)t 2) vf= vi + at 3) d= v(initial)t +1/2 acceleration times time^2 4)
ID: 1659693 • Letter: 1
Question
1) d=v(avg)t 2) vf= vi + at 3) d= v(initial)t +1/2 acceleration times time^2 4) v(final)^2 = v(initial)^2 + 2 acceleration times d I'm really struggling with my physics homework, I'm not looking for the answers, I have access to them. I just need help figuring out how to start these. 1) A ball rolls over the edge of a table with only a horizontal velocity. The height of the table is 1.3m and the ball hits the floor 2.1m away from the base of the table. With what velocity did the ball roll off the table? 2) A frustrated student throws their physics textbook horizontally with a velocity of 7.2 m/S from the bridge. It falls 19m to the water below. How far does the textbook travel horizontally before striking the water? 3) A sharpshooter points a rifle horizontally at the center of a target 45m away. The bullet leaves the barrel at 341 m/S. How far below the center of the target does the bullet hit?
Explanation / Answer
1)Height of table=h=1.3m
Horizontal distance travelled by ball while falling = 2.1m
Let time taken by ball to reach ground be t
h=(1/2)gt2
t = sqrt(2h/g)=sqrt(2*1.3/9.8)=0.515m
In 0.515 second, the horizontal distance travelled = 2.1m
Horizontal velocity = 2.1/0.515=4.08 m/s
2)Horizontal velocity=v=7.2 m/s2
Height by which the book falls = 19m
Time taken to fall by 19m = sqrt(2*19/9.8) = 1.969 seconds
Horizontal distance travelled in 1.969 seconds = 7.2*1.969=14.178 m
3) Horizontal velocity = 341 m/s
horizontal distance =45 m
time taken by bullet to reach target = 45/341=0.132seconds
The vertical height by which the bullet falls in 0.132 second = (1/2)*9.8*0.1322 = 0.0853 m = 8.53 cm
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