At the maximum height, all the kinetic energy is converted into potential energy. The gravitational potential energy at the maximum height is given by: \[ PE = mgh \]
where:
- \(m = 0.1 \, {kg}\) (mass),
- \(g = 10 \, {m/s}^2\) (acceleration due to gravity),
- \(h\) is the maximum height.
The initial kinetic energy \(KE\) is 20 J.
By conservation of energy:
\[ KE = PE \quad \Rightarrow \quad 20 = 0.1 \times 10 \times h \]
Solving for \(h\): \[ h = \frac{20}{1} = 20 \, {m} \]
Hence, the correct answer is (D).
From a height of 'h' above the ground, a ball is projected up at an angle \( 30^\circ \) with the horizontal. If the ball strikes the ground with a speed of 1.25 times its initial speed of \( 40 \ ms^{-1} \), the value of 'h' is: