Question

An object is thrown vertically up and attains an upward velocity of 34 m/s when it reaches one fourth of its maximum height above its launch point. What was the initial speed of the object?

Answer 1

Using kinematic equations and the given conditions, it has been determined that the initial speed of the object thrown vertically upwards was 68 m/s.

Step-by-step explanation:

To find the initial speed of the object thrown vertically upwards, we need to apply the principles of kinematic equations for uniformly accelerated motion. The problem states that the object attains a velocity of 34 m/s at one fourth of its maximum height. Since the only acceleration acting on the object is due to gravity, we will use the following equation which relates velocity, acceleration, and displacement:

v^2 = u^2 + 2as

Where:

• v is the final velocity (34 m/s)
• u is the initial velocity (what we are looking for)
• a is the acceleration due to gravity (-9.8 m/s^2)
• s is the displacement (1/4 of the maximum height H)

At the maximum height, the velocity is 0 m/s. Thus, we can apply the same equation to find the maximum height H:

0 = u^2 + 2(-9.8)(H) ---->(1)

Also, we know that at 1/4H the velocity is 34 m/s:

34^2 = u^2 + 2(-9.8)(H/4) ---->(2)

Dividing equation (2) by equation (1), we obtain:

(34^2) / (u^2) = (2(-9.8)(H/4)) / (2(-9.8)(H))

After simplifying, we find that u^2 is four times 34^2, or:

u^2 = 4 * 34^2

u = sqrt(4 * 34^2)

u = 2 * 34

u = 68 m/s

So, the initial velocity was 68 m/s.