# Given an initial launch velocity of 19m/s @ 31degrees, determine: The maximum height The time to reach the max height The total horizontal travel distance The total travel time What is the...

Given an initial launch velocity of 19m/s @ 31degrees, determine:

The maximum height

The time to reach the max height

The total horizontal travel distance

The total travel time

What is the horizontal velocity at any point

The vertical velocity after 1.5 seconds

The vertical position after 1.5 seconds

The horizontal position after 1.5 seconds

### 2 Answers | Add Yours

Initial velocity, u = 19 m/s

`theta` = 31 degrees

Using the equations of projectile motion:

Maximum height, H = `(u^2 sin^2theta)/(2g) = (19^2 sin^2(31))/(2xx9.81) = 4.88 m`

Total time of flight, t = `(2usintheta)/(g) = (2xx19xxsin31)/(9.81) = 2 sec`

Horizontal distance traveled or range = `(u^2sin2theta)/(g) = (19^2 sin(2xx31))/(9.81) = 32.49 m`

Time to reach maximum height = `(u sintheta)/(g) = (19xxsin31)/(9.81) = 1 sec`

Horizontal velocity at any point, ux = u cos 31 = 16.29 m/s

Vertical velocity after 1.5 sec = uy - gt = 19 sin 31 - 9.81x1.5 = **-4.93 m/s** (velocity is in downward direction)

Hope this helps.

So the vertical position after 1.5 seconds is 4.93meters downward? And the horizontal position is 0?