Question

a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/sec^2) find the maximum height

Answer 1

To find the maximum height, we need to use the initial velocity and the angle of the baseball's trajectory. By breaking down the initial velocity into its horizontal and vertical components, we can then use the equation h = v_y^2 / (2g) to find the maximum height.

Step-by-step explanation:

To find the maximum height, we first need to break down the initial velocity into its horizontal and vertical components. The initial velocity of the baseball is given as 100 ft/s at an angle of 45° with respect to the ground.

The horizontal component of velocity can be found using the equation: vx = v * cos(θ), where v is the initial velocity and θ is the angle.

The vertical component of velocity can be found using the equation: vy = v * sin(θ).

Once we have the vertical component of velocity, we can use the equation h = vy2 / (2g) to find the maximum height.
Substituting the given values, we have:
h = (100 * sin(45°))2 / (2 * 32)

Calculating this will give us the maximum height of the baseball above its initial position.

Answer 2

hmax=81ft

Step-by-step explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

`0=V_(y)-g*t=v_(0)*sin(\alpha)-g*th\\`

we isolate th (needed to reach the maximum height hmax)

`th = (v_(0)*sin(\alpha))/(g)`

The formula describing vertical distance is:

`y = Vy * t-g* t^(2) / 2`

So, given y = hmax and t = th, we can join those two equations together:

`hmax = Vy* th-g*th^(2)/2`

`hmax =Vo^(2)*sin(\alpha )^(2)/(2*g)`

if we launch a projectile from some initial height h all you need to do is add this initial elevation

`hmax =h+Vo^(2)*sin(\alpha)^(2)/(2*g)`

`hmax =3+100^(2)*sin(45)^(2)/(2 * 32)=81 ft`