A ball is thrown vertically upward with a speed of 19.0 m/s.
(a) How high does it rise?
(b) How long does it take to reach its highest point?
(c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is its velocity when it returns to the level from which it started?
v = 19 - 9.81 t
at top v = 0
t = 19/9.81 = 1.94 seconds to top
at top
h = 0 + 19 t - 4.9 t^2
h = 18.4 m high at top
the same as the rise time
1.94 s
the same 19 m/s but down now
To find the answers to these questions, we can use kinematic equations and formulas related to the motion of objects under constant acceleration.
(a) To determine how high the ball rises, we need to find the maximum height it reaches. We can use the following kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf is the final velocity (which is 0 when the ball reaches the maximum height),
vi is the initial velocity (19.0 m/s),
a is the acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2),
d is the displacement.
Rearranging the equation, we get:
d = (vf^2 - vi^2) / (2a)
Substituting the given values, we have:
d = (0 - 19.0^2) / (2 * -9.8)
Simplifying the equation, we find:
d ≈ 18.86 meters
So, the ball rises approximately 18.86 meters.
(b) The time it takes to reach the highest point can be determined using the following kinematic equation:
vf = vi + at
Where:
vf is the final velocity (which is 0 when the ball reaches the maximum height),
vi is the initial velocity (19.0 m/s),
a is the acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2),
t is the time.
Rearranging the equation, we get:
t = (vf - vi) / a
Substituting the given values, we have:
t = (0 - 19.0) / -9.8
Simplifying the equation, we find:
t ≈ 1.95 seconds
So, it takes approximately 1.95 seconds to reach the highest point.
(c) The time it takes for the ball to hit the ground after reaching its highest point is equal to the time it took to reach the highest point, since the motion follows the same path on the way down. So, it also takes approximately 1.95 seconds for the ball to hit the ground.
(d) The velocity when the ball returns to the level from which it started can be determined using the same kinematic equation used in part (b):
vf = vi + at
Substituting the given values, we have:
vf = 0 + (-9.8) * 1.95
Simplifying the equation, we find:
vf ≈ -19.1 m/s
The negative sign indicates that the velocity is directed downwards.
So, the velocity when the ball returns to the level from which it started is approximately -19.1 m/s.