It is still going 1.61 m/s horizontal if it was going that speed originally.
Vertical the speed is:
v = Vo + a t, the initial speed down, Vo, is zero
v = 0 - 9.8 t
what is t, the time to fall 1.3 m?
h = ho + Vo t + .5 a t^2
-1.3 = 0 + 0 - 4.9 t^2
t = .515 s
so
v = -9.8 * .515 = -5.05 m/s
speed = sqrt (-5.05^2 + 1.61^2)
= sqrt ( 25.5+2.59)
=5.3 m/s
tan angle to horizontal = 5.05/1.61 = 3.14
angle down from horizontal = 72.3 degrees down from horizontal
Could u help with physics hw! i keep trying but i keep getting the wrong answer?
a 71g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of .83m away from the table.
1.how fast was it rolling on the table before it fell off?
ok well i already figured out the first question. which was 1.61
but for some reason i keep getting the
2nd question wrong.
can someone help!please
2.What was the ball's velocity just before it hit the floor?
That is, at what angle in the range -90° to +90° relative to the horizontal directed away from the table) did the ball hit the floor?
answer in units of °
1 answer
2 answers