A stone of mass 0.7 kg is projected vertically upward with a speed of 5 $ms^{-1}$. Calculate the maximum height reached. [g =10$ms^{-2}$]

Analyze the question

A stone is thrown straight up with an initial speed of 5 m/s. At its maximum height, its velocity becomes zero. We need to find that height. The mass (0.7 kg) is extra information, because the height does not depend on mass when air resistance is ignored.

Concept

For motion under gravity, use the equation:

v² = u² − 2gh

where

• v = final velocity (0 at the top)
• u = initial velocity = 5 m/s
• g = 10 m/s²
• h = maximum height

Solution

0 = 5² − 2(10)h

0 = 25 − 20h

20h = 25

h = 25/20 = 1.25 m

Check with energy: The kinetic energy at launch turns into potential energy at the top.

½mu² = mgh, so h = u²/2g = 25/20 = 1.25 m. The mass cancels, which confirms it is not needed.

Answer: A. 1.25 m

Tip: For maximum height problems, remember h = u²/2g. If g = 10, just square the speed and divide by 20. Also watch out for extra data like mass, which is often placed in the question to distract you.