Question

A projectile is shot in the air from ground level with an initial velocity of 560 m/sec at an angle of 30° with the horizontal. (Round your answers to two decimal places.) (a) At what time (in seconds) is the maximum range of the projectile attained? s (b) What is the maximum range (in meters)?

Answer 1

a) t = 57.14 s

b) x = 27711.4 m

Step-by-step explanation:

This is a missile throwing exercise

a) They ask us for the time to the maximum reach, this corresponds to when it reaches the ground y = 0, let's use

y =
`v_(oy)` t - ½ g t²

Let's use trigonometry to find the vertical initial velocity

sin θ = v_{oy} / v₀

v_{oy} = v₀ sin θ

we substitute

y = v₀ sin θ t - ½ g t²

since the height is zero

0 = t (v₀ sin θ - ½ g t)

This equation has two solutions

* t = 0 which corresponds to the moment of launch

*

v₀ sin 30 - ½ g t = 0

t = v₀ sin 30 2/g

let's calculate

t = 560 sin 30 2 / 9.8

t = 57.14 s

b) with this time we can calculate the distance traveled

x = v₀ₓ t

let's use trigonometry for velocity

cos θ = v₀ₓ / v₀

v₀ₓ = v₀ cos 30

we substitute

x = v₀ cos 30 t

let's calculate

x = 560 cos 30 57.14

x = 27711.4 m