Q23. Object is thrown vertically upwards and has a speed of 18 m/s when it reaches one-fourth of its maximum height h above its launch point. Label the relevant points and answer the following;
a) Determine the maximum height h.
b) What is the initial speed of throwing the object?
at h/4
total mechanical energy
= m g(h/4)+(1/2)m (18)^2
at h
total mechanical energy
= m g h + 0
so
gh = gh/4 + 9*18
for part b
(1/2)m v^2 = m g h
To answer part (a), we can use the concept of motion under constant acceleration. We know that the object reaches one-fourth of its maximum height with a speed of 18 m/s. Let's label the relevant points as follows:
Launch point: A
One-fourth of maximum height: B
Maximum height: C
At point A, the initial speed is v0 (the initial speed of throwing the object) and the height is 0 m.
At point B, the speed is 18 m/s and the height is one-fourth of the maximum height h.
At point C, the speed is 0 m/s and the height is the maximum height h.
Now let's solve the problem step by step:
Step 1: Find the time taken to reach point B.
To find the time taken to reach point B, we can use the equation of motion for vertical motion:
v = v0 + at,
where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.
Since the object is moving vertically upwards, the acceleration is -9.8 m/s^2 (taking downward direction as negative).
At point B, v = 18 m/s, v0 = initial velocity (unknown), a = -9.8 m/s^2, and t = time taken to reach point B (unknown). Plugging in these values, we get:
18 = v0 + (-9.8)t.
Step 2: Find the time taken to reach point C.
At point C, the object reaches its maximum height. At this point, the final velocity is 0 m/s. We can use the same equation of motion as in step 1 to find the time taken to reach point C:
0 = v0 + (-9.8)t2.
Step 3: Find the relationship between the times taken to reach points B and C.
Since the object is thrown vertically upwards, the time taken to reach a certain height while going up is the same as the time taken to come back down to that same height. Therefore, the time taken from point A to point B is the same as the time taken from point B to point C.
Step 4: Use the relationship obtained in step 3 to solve for the maximum height h.
Let's assume the time taken to reach point B and point C is t. Using the relationship obtained in step 3, we can equate the two times:
t = t2.
Now, we have two equations:
18 = v0 + (-9.8)t,
0 = v0 + (-9.8)t.
We can solve these two equations simultaneously to find the initial speed (v0) and the time (t). Once we have the time (t), we can find the maximum height h using the equation of motion:
h = v0t + 0.5at^2.
Now let's move on to part (b) of the question.
To find the initial speed of throwing the object (v0), we need to solve the two equations derived in step 4, which are:
18 = v0 + (-9.8)t,
0 = v0 + (-9.8)t.
By solving these equations simultaneously, we can find the initial speed v0.