Question

1. An object was launched at 30⁰ angle and travel 55.75m. Assume That the object was launched on a level ground.

a. What was the take-off speed of the object?

Answer 1

Find the horizontal component of velocity:

$$v_x=v\text{ cos}30°.$$

55.75 m is the horizontal range. The time it takes to travel this distance at $v_x$ is

$$t=\frac{R}{v_x}=\frac{R}{v\text{ cos}30°}$$

On the other hand, this time is equal to the time required for the body to reach the highest point and then fall to the ground:

$$t=2\cdot\frac{v_y}{g}=\frac{2v\text{ sin}30°}{g}.$$

So, we have two equal time variables expressed differently.

$$\frac{2v\text{ sin}30°}{g}=\frac{R}{v\text{ cos}30°},\\\space\\ v=\sqrt{\frac{gR}{\text{sin}(2\cdot30°)}}=25.12\text{ m/s}.$$