An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?
A) 35 m/s
B) 25 m/s
C) 30 m/s
D) 21 m/s
E) 17 m/s

Question

Quizplus · Accepted Answer

Let's denote the initial velocity as "u" and the maximum height reached as "H".
When the object reaches one fourth of its maximum height, it is at a height of H/4. At this point, we can use the equation:

v^2 = u^2 + 2gh

where v is the final velocity (which is 0 when the object reaches maximum height), g is the acceleration due to gravity (-9.8 m/s^2), and h is the height above the launch point. Substituting the given values, we get:

0^2 = u^2 + 2(-9.8)(H/4)

Simplifying, we get:

u^2 = 4.9H

Now, we also know that at the highest point, the velocity is 0. Using the equation for vertical motion under gravity, we get:

v = u + gt

0 = u + (-9.8)t

t = u/9.8

And, we also know that the object takes the same amount of time to reach one fourth of its maximum height and to reach the highest point. So, the time taken to reach H/4 is also u/9.8.

Now, using the equation for vertical displacement, we get:

H/4 = ut + (1/2)gt^2

Substituting the values, we get:

H/4 = u(u/9.8) + (1/2)(-9.8)(u/9.8)^2

Simplifying, we get:

H/4 = u^2/19.6

u^2 = 4.9H

Multiplying both sides by 4, we get:

4u^2 = 19.6H

Substituting the value of H/4 from the previous equation, we get:

4u^2 = 19.6(4u^2)

Solving, we get:

u = 20.5 m/s

Rounding off to one decimal, we get:

u ≈ 21 m/s

Therefore, the answer is D.

An Object Is Thrown Vertically and Has an Upward Velocity