Given data

• Mass of ball: $m = 0.15\,\text{kg}$
• Maximum height reached after leaving machine: $h = 20\,\text{m}$
• Distance over which machine pushes ball: $s = 0.2\,\text{m}$
• Gravitational acceleration: $g = 10\,\text{m s}^{-2}$

Find the speed of the ball just after leaving the machine

At the highest point, velocity becomes zero. Using energy conservation from launch point to maximum height:

$\frac{1}{2}mv^2 = mgh$

$\frac{1}{2}v^2 = gh$

$v^2 = 2gh = 2 \times 10 \times 20 = 400$

$v = 20\,\text{m/s}$

Apply work-energy theorem during launching

While the machine pushes the ball through distance $s = 0.2\,\text{m}$, two forces do work on the ball:

• Upward force by machine: $F$
• Downward weight: $mg$

Net work done on the ball equals change in kinetic energy:

$Fs - mgs = \frac{1}{2}mv^2$

Substitute values:

$F(0.2) - (0.15)(10)(0.2) = \frac{1}{2}(0.15)(400)$

$0.2F - 0.3 = 30$

$0.2F = 30.3$

$F = \frac{30.3}{0.2} = 151.5\,\text{N}$

Integer answer

Since this is an integer-type question, the expected value is taken as:

$F \approx 152\,\text{N}$

Comparison with stored answer

The physically correct calculation gives $151.5\,\text{N}$, which rounds to $152\,\text{N}$, not $150\,\text{N}$.

If one neglects the small work done against gravity during the $0.2\,\text{m}$ push, then:

$Fs = \frac{1}{2}mv^2 = 30$

$F = \frac{30}{0.2} = 150\,\text{N}$

So the stored answer likely ignores gravity during contact.