Answer:
my bad i wish i could ansrew
Explanation:im only six
Answer:
answer is 0.1428
Explanation:
Data:- vf=5.0 , vi=0.0 , t=35 , a=? so appling first eq of motion vf=vi+at we have to find a=vf-vi/t , a=5.0-0.0/35 , a=5/35 ,a=0.1428m/sec²
The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.
m mass
v velocity
h height
g = 9.81m/s²
The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)
Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
