The height attained by the ball is 11.86m
a ball is shot from the ground straight up into the air its initial and final velocity is
initial velocity, u = 50 ft/s = 50×0.305 = 15.25m/s
final velocity ,v = 0 m/s
gravity =-9.8 m/s²
( negative sign shows acceleration in opposite direction)
height =?
using the newton motion of equation
v² = u² + 2as
where
a= acceleration due to gravity(g)
s = height
v² = u² + 2gs
(0)² = (15.25)² + 2×(-9.8)×s
0 = (15.25)² - 19.6 × s
s= - (15.25)²/ 19.6
s = 11.86m
after ignoring the air resistance the maximum height of the ball is 11.86m
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Answer:
can you explain more or..,.
Explanation:
Answer:
They slow down.
Explanation:
The collided so the slow down till they stop.
Answer:
W = 0.678 rad/s
Explanation:
Using the conservation of energy:

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.
First, we will find the moment of inertia as:
I =
where m is the mass and R the radius, so:
I =
I = 36.26 Kg*m^2
Then, replacing values on the initial equation, we get:

also we know that:
V =WR
so:

Finally, solving for W, we get:

W = 0.678 rad/s