Question

Consider a projectile launched with an initial velocity of v0 at an angle of θ . Assume acceleration due to gravity is g.

What will be the maximum (vertical) height of the projectile (assuming a general angle, θ )?

Answer 1

A.

Step-by-step explanation:

`1. the \ initial \ formula \ is\\ \ mgh=(m(Vsin \theta)^2)/(2); \\ 2. \ finally, \\ h=(V_0^2sin^2 \theta)/(2g).`