Question

A projectile is launched with an initial velocity of 25.0 m/s at an angle of 60o above the horizontal. At some time after its launch, the projectile’s velocity vector makes an angle of 30o with respect to the horizontal. What is the projectile’s speed at that point?

Answer 1

14.4 m/s

Step-by-step explanation:

`v_(o)` = initial velocity of the projectile at the time of launch = 25 m/s

`\theta _(o)` = initial angle of launch = 60°

`v_(ox)` = initial velocity of the projectile in horizontal direction=
`v_(o) Cos\theta _(o)` = 25 Cos60 = 12.5 m/s

`v_(f)` = final velocity of the projectile =
`v`

`\theta _(f)` = final angle = 30°

`v_(fx)` = final velocity of the projectile in horizontal direction=
`v_(f) Cos\theta _(f)` =
`v` Cos30

we know that the velocity along the horizontal direction remains constant, hence

`v_(fx)` =
`v_(ox)`

`v` Cos30 = 12.5

`v` = 14.4 m/s