Answer to Question #194798 in Physics for Maria Natasha

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:

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