Question

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

Answer 1

Answer:

20.78 m/s that we can approximate to option d (21 m/s)

Explanation:

The solution involves a lot of algebra and to be familiar with different convenient formulas for launching an object vertically under the action of gravity.

First you need to recall (or derive) the formula for the maximum height reached by an object with launches with initial velocity $v_0$:

Maximum height = $\frac{{v_0}^2}{2g}$

Therefore one fourth of such height would be: $\frac{{v_0}^2}{8g}$

Second, find what would be the time needed to reach that height by solving for the time in the equation for the vertical position:

$y(t)=v_0*t-\frac{g}{2} t^2 \\\frac{{v_0}^2}{8g} = v_0*t-\frac{g}{2} t^2\\\frac{g}{2} t^2-v_0*t+\frac{{v_0}^2}{8g}=0$

And now, solve for t in the last equation using the quadratic formula to find the time needed for the object to reach that height (one fourth of the max height):

$t=\frac{v_0+/-\sqrt{{v_o}^2-4*\frac{g}{2}*\frac{{v_o}^2}{8g}  } }{g} = \frac{v_0+/-\sqrt{{v_o}^2-\frac{{v_o}^2}{4}  } }{g} =\frac{v_0+/-\sqrt{\frac{3{v_o}^2}{4}  } }{g} = \\=\frac{v_0+/-v_0\sqrt{\frac{3}{4}  } }{g} =\frac{v_0+/-v_0\frac{\sqrt{3}}{2}  } {g}$

Next, use this expression for t in the equation for the velocity at any time t in the object's trajectory that comes from the definition of acceleration;

$v(t)=v_0-g*t$

Then for the time we just found, this new equation becomes:

$v=v_0-g(\frac{v_0+/-v_0\frac{\sqrt{3} }{2}}{g}) =v_0-v_0+/- v_0\frac{\sqrt{3} }{2} = +/- v_0 \frac{\sqrt{3} }{2}$

Now, using that the velocity at this height is 18 m/s, and solving for the unknown velocity $v_0$, we get:

$v_0=\frac{18*2}{\sqrt{3} } =\frac{36}{\sqrt{3} }= 20.78\frac{m}{s}$

Answer 2

The initial (launch) speed of the object is 21 m/s.

Initial speed of the projectile

The initial speed of the projectile is calculated by using the following formulas as shown below;

v² = u² - 2gh

u² = v² + 2gh

u² = (18)² + 2(9.8)(0.25h)

u² = 324 + 4.9h  ---- (1)

When the object reaches maximum height

At maximum height, the final velocity of the object = 0

u² = v² + 2gh

u² = 0 + 2gh

u² = 2gh

u² = 2(9.8)(h)

u² = 19.6h    ---- (2)

solve (1) and (2) together

324 + 4.9h = 19.6h

324 = 14.7h

h = 324/14.7

h = 22.04 m

Magnitude of the initial speed

u² = 19.6 x 22.04

u² = 431.98

u = √431.98

u = 20.8 m/s

u ≅ 21 m/s

Thus, the initial (launch) speed of the object is 21 m/s.

Learn more about initial speed here: https://brainly.com/question/24493758