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?

a)   = 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

Explanation:

This exercise is a projectile launch

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

          = 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 =

         t =

         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 = -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 =

          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

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min