Question

An object is launched at 45ᵒ to the horizontal on the level ground. What is the range of the projectile if it has an initial velocity of 180 ft/s? Neglect air resistance.

Answer 1

We can find the range of the projectile from the formula:

$$R=\dfrac{v^2sin2\theta}{g},$$ $$R=\dfrac{(180\ \dfrac{ft}{s})^2sin2\cdot45^{\circ}}{32.17\ \dfrac{ft}{s^2}}=1007\ ft.$$