Question

A projectile is fired into the air at a nonzero angle with the horizontal. When the projectile reaches its maximum height, the speed of the projectile is 50.8% of its original speed. What angle was the projectile originally fired at?

Answer 1

59.5 deg

Step-by-step explanation:

`v` = original speed at which the projectile is launched

θ = angle of launch of projectile

`v_(x)` = component of speed along the horizontal direction =
`v` Cosθ

At the highest position, the vertical component of velocity becomes zero and there is only horizontal component of velocity, hence

`v_(highest)` = velocity at the highest point =
`v_(x)` =
`v` Cosθ

it is given that

`v_(highest)` = 0.508
`v`

so

`v` Cosθ = 0.508
`v`

Cosθ = 0.508

θ = 59.5 deg