Question

A ball is thrown vertically upward. It has a speed of \(10\ m/s\) when it has reached one half of its maximum height. How high does the ball rise?
(Take \(g = 10\, m/s^2\))

• A. 10 m
• B. 5 m
• C. 15 m
• D. 20 m

Answer 1

The Correct Option is A

To solve this problem, we need to determine the maximum height reached by the ball. We know from the problem that the ball has a speed of \(10\ m/s\) at half of its maximum height. Let's denote:

• v = final velocity at maximum height = 0\ m/s (since the ball stops for a moment at its peak)
• u\) = initial speed at half of its maximum height = 10\ m/s\)
• g =\, -10\ m/s^2\) (negative because gravity decelerates the ball)
• h\) = maximum height
• h/2\) = half of its maximum height

Using the first equation of motion between half height and maximum height:

v^2 = u^2 + 2gh\)

Substitute v = 0\), u = 10\ m/s\), and gh\) = -5\ g\):

0 = (10)^2 + 2 \times (-10) \times h/2\)

Simplify:

0 = 100 - 10h\)

10h = 100\)

h = 10\ m\)

Thus, the maximum height that the ball rises is 10 m.