Question

A projectile is shot horizontally from the edge of a cliff, 400 m above the ground. The projectile is traveling horizontally at 65 m/s the moment it leaves the cannon.

What is the initial vertical velocity of the projectile?

Determine the time taken by the projectile to hit the ground.

How far does the projectile land from the base of cliff?

Find the vertical component of the velocity just before the projectile hits the ground.

What is the magnitude of the final vector velocity the instant before it hits the ground?

What is the direction of the final vector velocity the instant before it hits the ground?

Answer 1

a)
`v_(oy)` = 0, b) t = 9.035 s, c) x = 587.275 m, d) v_y = -88.54 m / s,

e) v = 109.84 m / s, f) θ = -53.7º

Step-by-step explanation:

This exercise is a projectile launch

a) as the projectile is launched horizontally, its initial vertical velocity is zero

`v_(oy)` = 0

b) let's use the relation

y = y₀ + v_{oy} t - ½ g t²

when reaching the ground y = 0 and the initial vertical velocity is v_{oy}=0

0 = y₀ + 0 - ½ g t²

t =
`\sqrt{ (2 v_o)/(g) }`

t =
`\sqrt{ ( 2 \ 400 )/( 9.8) }`

t = 9.035 s

c) the horizontal distance is

x = v₀ₓ t

x = 65 9.035

x = 587.275 m

d) the vertical speed when touching the ground

v_y =
`v_(oy)` -gt

v_y = 0 - 9.8 9.035

v_y = -88.54 m / s

the negative sign indicates that the velocity is directed downwards

e) the velocity at this point

v =
`√( v_x^2 + v_y^2)`

v =
`√(65^2 + 88.54^2)`

v = 109.84 m / s

f) the direction of the velocity

let's use trigonometry

tan θ = v_y / vₓ

θ = tan⁻¹1 v_y / vₓ

θ = tan⁻¹ (-88.54 / 65)

θ = -53.7º

this angle is measured clockwise