Question

A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reaches above its starting point in terms of v0, θ0, and g. (Ignore any effects due to air resistance.)

Answer 1

The maximum height (h) a projectile reaches is given by the formula h = v0^2 sin^2(θ0) / (2g), where v0 is the initial velocity, θ0 is the launch angle, and g is the acceleration due to gravity.

Step-by-step explanation:

To find the expression for the maximum height reached by a projectile launched at an angle θ0 with an initial speed v0, we start by considering the vertical component of the initial velocity, which is v0y = v0 sin(θ0). The maximum height (h) is achieved when the vertical velocity becomes zero (vy = 0), which occurs at the peak of the trajectory. Using kinematic equations, we know that the final velocity squared is equal to the initial velocity squared minus two times the acceleration times the height (v^2 = v0^2 - 2gh). Substituting vy for v and v0y for v0, the equation becomes 0 = v0^2 sin^2(θ0) - 2gh. Solving for height (h), we get:

h = v0^2 sin^2(θ0) / (2g)

This equation shows that the maximum height is directly proportional to the square of the initial velocity's vertical component and inversely proportional to twice the acceleration due to gravity.

Answer 2

`h= ((v_(o))^(2) sin^(2) \theta o )/(2g)`

Step-by-step explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t Equation (1)

Where:

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀ : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y²
`h= ((v_(o))^(2)sin^(2) \theta o )/(2g)`

2gh= (v₀*sin θ₀)²

`h= (v_(o)^(2) sin^(2) \theta o )/(2g)`