A ball is projected vertically upward at a speed of 50 m/s . Determine the speed at half the maximum height. Take g = 10 m/s^2 .

Correct answer: B. 25 2 m/s Initial speed u = 50 m/s Maximum height H = (u^2)/(2g) Let v be the speed at half the maximum height, h = (H)/(2) . Using the equation of motion: v^2 = u^2 - 2gh v^2 = u^2 - 2g ((H)/(2)) v^2 = u² - gH Substituting H = (u^2)/(2g): v^2 = u² - g ((u^2)/(2g)) = u^2 - (u^2)/(2) = (u^2)/(2) v = u 2 v = 50 2 ==25 2 m/s

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Rest and Motion Kinematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rest and Motion Kinematics

A ball is projected vertically upward at a speed of 50 m/s ....

A ball is projected vertically upward at a speed of

50 \text{ m/s}

. Determine the speed at half the maximum height. Take

g = 10 \text{ m/s}^2

Options

A $50\sqrt{2} \text{ m/s}$
B $25\sqrt{2} \text{ m/s}$
C $35 \text{ m/s}$
D $25 \text{ m/s}$

Correct answer

B. $25\sqrt{2} \text{ m/s}$

Step-by-step solution

Initial speed

u = 50 \text{ m/s}

Maximum height

H = \dfrac{u^2}{2g}

Let

v

be the speed at half the maximum height,

h = \dfrac{H}{2}

.

Using the equation of motion:

v^2 = u^2 - 2gh

v^2 = u^2 - 2g

\left(\dfrac{H}{2}\right)

v^2 = u² - gH

Substituting

H =

\dfrac{u^2}{2g}:

v^2 = u² - g \left(\dfrac{u^2}{2g}\right) =

u^2 -

\dfrac{u^2}{2} = \dfrac{u^2}{2}<br><br>

v =

\dfrac{u}{\sqrt{2}}<br><br>

v = \dfrac{50}{\sqrt{2}} ==25\sqrt{2} \text{ m/s}

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