Question

A projectile is launched with speed

v₀ and angle θ. Derive an expression for the projectile's maximum height h.
Express your answer in terms of the variables v₀, θ, and appropriate constants

Answer 1

H = V0y t - 1/2 g t^2 since V0 and g have different directions

t = V0y / 9.8 m/s^2 time for projectile to lose its vertical speed

V0y = V0 sin θ initial vertical speed

t = V0 sin θ / 9.80

H = V0 sin θ * V0 sin θ / 9.80 - 1/2 g * (V0 sin θ / 9.80)^2

H = V0^2 / 9.80 * sin^2 θ - 1/2 g * (V0 sin θ / 9.80)^2

Answer 2

The maximum height of a projectile launched with speed v₀ and angle θ can be derived using the kinematic equations for vertical motion. The expression for the maximum height is given by (v₀²sin²(θ)) / 2g, where g is the acceleration due to gravity.

Step-by-step explanation:

The maximum height h of a projectile launched with speed v₀ and angle θ can be derived by considering the vertical component of the initial velocity. The vertical component of the initial velocity is given by v₀sin(θ) and the acceleration due to gravity is g. Using the kinematic equation for vertical motion, we can find the time it takes for the projectile to reach its maximum height:

(vf - vy) = -gt
0 - v₀sin(θ) = -gt
t = v₀sin(θ) / g

Substituting this time into another kinematic equation, we can find the maximum height:

h = vy * t + 0.5 * (-g) * t²
h = (v₀sin(θ))(v₀sin(θ) / g) + 0.5 * (-g) * (v₀sin(θ) / g)²
h = (v₀²sin²(θ)) / 2g

Therefore, the expression for the maximum height h of the projectile is (v₀²sin²(θ)) / 2g, where g is the acceleration due to gravity.