Question

A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.

1. Calculate the time the projectile takes to hit the ground
2. Calculate the distance the projectile travelled on the X-axis horizontally before hitting the ground
3. Calculate the impact velocity of the projectile on the Y-axis
4. Calculate the Resultant velocity V r
5. Calculate the angle at which the projectile hits the ground
6. Calculate Sine Θ
7. Calculate Tan Θ
8. A final Resultant Velocity Statement

Answer 1

1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Step-by-step explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

y = y₀ +
`v_(oy)` t - ½ g t²

When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0

0 = y₀ - ½ g t2

t = √ 2y₀ / g

t = √ (2 60 / 9.8)

t = 3.45 s

2) The horizontal distance traveled

x = v₀ₓ t

x = 40 3.45

x = 138 m

3) The vertical velocity at the point of impact

v_{y} = I go - g t

v_{y} = 0 - 9.8 3.45

v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

v = √ (vₓ² + v_{y}²)

v = √ (40² + 33.8²)

v = 52.37 m / s

5) angle of impact

tan θ = v_{y} / vx

θ = tan⁻¹ v_{y} / vx

θ = tan⁻¹ (-33.81 / 40)

θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis