Question

a projectile is fired from the surface of hte earth with a speed of 200 meters per second at a an angle of 30 degrees above the horizontal. if the ground is level what is the max height reached by the projectile

Answer 1

506.68 meters

Step-by-step explanation:

When you launch a projectile at an angle θ from the horizontal, the initial velocity of the projectile will have a vertical and a horizontal component.

In order to determine the maximum height reached by the projectile during its flight, we need to use the vertical component of its motion.

We can use the equation below to find the maximum height.

`h_(max) =(v_o^(2)sin^2(B) )/(2g)`

We are given

`v_o=200`

`B=30`

`g=9.81`

Lets evaluate
`h_(max)`.

`h_(max) =(200^(2)sin^2(30) )/(2*9.81)`

`h_(max) =(40000^(2)sin^2(30) )/(2*9.81)`

`h_(max) =(40000((1)/(2))^2 )/(2*9.81)`

`h_(max) =(40000((1)/(4)) )/(2*9.81)`

`h_(max) =(10000)/(2*9.81)`

`h_(max) =(10000)/(19.62)`

`h_(max) =509.68`