A ball is thrown upward in such a way that its speed is 32.3 m/s when it is at half its maximum height. Find
(a) its maximum height,
(b) its velocity 2.0 s after it's thrown,
(c) its height 2.0 s after it's thrown, and
(d) its acceleration at its maximum height.
Please show work.
(a) v=vₒ-gt
v=0
t=vₒ/g.
h= vₒt-gt²/2=vₒ²/2g =32.3²/2•9.8=53.23 m.
(b) v= vₒ - g•t1 =
= 32.3 - 9.8•2=12.7 m/s
(c) h= vₒ•t1-g•t²/2=
=32.3•2 – 9.8•4/2 =42 m
(d) g=9.8 m/s² (eveywhere)
To solve this problem, we can use the equations of motion for vertical motion under constant acceleration. The key equation we need is:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Given information:
Initial velocity, u = 32.3 m/s (when the ball is at half its maximum height)
(a) To find the maximum height:
At the maximum height, the final velocity is 0 m/s. Thus, using the equation above, we can write:
0 = u^2 + 2as
Rearranging this equation, we get:
as = -u^2
The negative sign indicates that the acceleration is in the direction opposite to the initial velocity. We also have the equation for displacement:
s = ut + (1/2)at^2
Since the final velocity (v) is 0 m/s at maximum height, this equation simplifies to:
s = ut
Substituting the values, we get:
s = 32.3 * t
where t is the time taken to reach maximum height.
Now, let's move on to part (b) and (c) of the question:
(b) To find the velocity 2.0 s after the ball is thrown:
We can use the equation:
v = u + at
Given that the initial velocity, u, is 32.3 m/s and the time, t, is 2.0 s, we can calculate the acceleration using the equation above.
(c) To find the height 2.0 s after the ball is thrown:
We can use the equation for displacement:
s = ut + (1/2)at^2
Given that the initial velocity, u, is 32.3 m/s and the time, t, is 2.0 s, we can calculate the displacement.
Now, let's move on to part (d) of the question:
(d) To find the acceleration at the maximum height:
From the equation as = -u^2, we have already determined the product of acceleration and displacement at the maximum height. We can use this to find the acceleration.
By solving these equations one by one, we can find the answers to the questions.