Question

Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reaches the maximum height of 15 m. The acceleration due to gravity is 30 m/s2. Find the horizontal range of the projectile in meters.

Answer 1

0.0503 = r

Step-by-step explanation:

Answer 2

R = 0.0503 m

Step-by-step explanation:

This is a projectile launching exercise, to find the range we can use the equation

R = v₀² sin 2θ / g

How we know the maximum height

`v_(f)`² =
`v_(oy)`² - 2 g y

`v_(f)`= 0

`v_(oy)` = √ 2 g y

`v_(oy)` = √ 2 9.8 / 15

`v_(oy)` = 1.14 m / s

Let's use trigonometry to find the speed

sin θ =
`v_(oy)` / vo

vo =
`v_(oy)` / sin θ

vo = 1.14 / sin 60

vo = 1.32 m / s

We calculate the range with the first equation

R = 1.32² sin(2 60) / 30

R = 0.0503 m