Question

A projectile is launched from ground level with an initial speed of 47.5 m/s at an angle of 33.3° above the horizontal. It strikes a target in the air 2.82 seconds later. What are the horizontal and vertical distances from where the projectile was launched to where it hits the target?

Answer 1

Answer on Question #63013, Physics / Mechanics | Relativity

A projectile is launched from ground level with an initial speed of 47.5 m/s at an angle of 33.3° above the horizontal. It strikes a target in the air 2.82 seconds later. What are the horizontal and vertical distances from where the projectile was launched to where it hits the target?

Solution:

$$x = v_{0x} * t = v_0 \cos \theta * t$$ $$x = 47.5 * \cos(33.3) * 2.82 = 112.52 \, \text{m}$$ $$y_0 = v_{0y} * t - \frac{1}{2} g t^2 = v_0 \sin \theta * t - \frac{1}{2} g t^2$$ $$y = 47.5 * \sin(33.3) * 2.82 - 0.5 * 9.8 * 2.82^2 = 88.46 \, \text{m}$$

Answer: 112.52 m and 88.46 m

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