Question

A projectile is launched at 60° from the horizontal and lands 50m from the launch point. What was the projectile's speed at launch?

Answer 1

Answer on Question #83034, Physics / Molecular Physics | Thermodynamics

**Question:**

A projectile is launched at $60{}^{\circ}$ from the horizontal and lands $50\mathrm{m}$ from the launch point. What was the projectile's speed at launch?

**Solution:**

The distance $L$ from the launch point to the landing point equals to

$$L = v \cos \alpha \cdot T = v \cos \alpha \cdot 2 \frac {v \sin \alpha}{g} = \frac {v ^ {2} \sin 2 \alpha}{g}, \text{ respectively } v = \sqrt {\frac {L g}{\sin 120{}^{\circ}}} = \sqrt {\frac {50 \cdot 10 \cdot 2}{\sqrt {3}}} = 24$$

(m/s).

**The answer:**

$$v = \sqrt {\frac {L g}{\sin 120{}^{\circ}}} = \sqrt {\frac {50 \cdot 10 \cdot 2}{\sqrt {3}}} = 24 \ \mathrm{m/s}.$$

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