# Calculating Mass for a Reduced Trajectory while Keeping Projectile Force Constant

Apologies if this is not a research-specific question, but applied mechanics.

A firefighter has to pass a test in which he is to throw a medicine ball a certain distance. The ball is thrown at a 45 degree angle (Fig. 1) and has to travel a ground distance of 7.5 meters. For the test, a medicine ball is a rough sphere with a mass of 4 kg and ~250 mm in diameter. In preparing for the test, however, the firefighter has only access to a dumbbell, which he does not intend to throw.

Given that a certain force is needed to throw the medicine ball with a preparation phase of 1 m and then released for a 7.5 meter throw, how heavy should the training substitute dumbbell be in order to achieve the equivalent force? Assume that the dumbbell only covers a 1 m distance at 45 degrees and that the firefighter intends to train at a 1 m/s velocity.

Figure 1

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## 1 Answer

For a projectile the range is given by $$R=\frac{u^2\sin2\theta}{g}$$ Where $u$ is the initial speed of projectile and $\theta$ is the angle of projection.
Substituting $R=7.5m, g=10 m/{s^2}, \theta = \frac \pi4$, we get $u$ as $$u=\sqrt{gR}=\sqrt{7.5 * 10} = 5\sqrt3 m/s$$ So the change in momentum of the ball is $$\Delta P = mu-0$$ $$\therefore \Delta P = 20\sqrt3 kgm/s$$ Change in momentum equals impulse applied by the thrower. Thus during training he must also practise to apply the same impulse. Change in momentum of the dumbbell will be $$\Delta P = m_{dumbbell}*1m/s$$ $$\therefore 1*m=20\sqrt3$$ $$m=20\sqrt3kg$$.
Note: The equations of projectile assume there is no air resistance, and that the projectile is being launched from ground level.(A formula can be derived considering it is thrown from an initial height).

Also, during the practise, there will be a lot of other factors involved like decelerating the dumbbell as you don't intend to throw it. I have ignored all those effects.

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Thanks! 34 kg sounds rather heavy to me. If I understand correctly, this would correspond to approximately 17 kg at 2 m/s (500ms/m) or 9 kg at 4 m/s (250ms/m). –  noumenal Jul 11 '13 at 17:09
yes you are correct @noumenal –  udiboy1209 Jul 11 '13 at 17:57