Question

A projectile is launched at ground level with an initial speed of 50.0 m/s and an
elevation angle (i.e., at an angle above the horizontal) of 30.0°. It strikes a target above the
ground 3.00 s later. What are the horizontal and vertical distances of the target from the place
where the projectile was launched? Ignore air resistance.

Answer 1

Answer on Question #62297-Physics -Mechanics

A projectile is launched at ground level with an initial speed of 50.0 m/s and an elevation angle (i.e., at an angle above the horizontal) of 30.0°. It strikes a target above the ground 3.00 s later. What are the horizontal and vertical distances of the target from the place where the projectile was launched? Ignore air resistance.

Solution

$$x = v \cos 30 t = 50 \cos 30 3 = 130 \, \text{m}.$$ $$y = v \sin 30 t - \frac{g t^2}{2} = 50 \sin 30 3 - \frac{9.81 \cdot 3^2}{2} = 30.9 \, \text{m}.$$

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