Aprojectline is shot from the edge of a cliff 100 m above the ground with intial speed 10 m/s at an angle 30 with the horizontal.

1) determine the time taken by the projectile motion to hit the ground

2) determine the range measured from the base of the cliff

3)vertical component of the velocity just before it hit the ground

4) ,agnitude to the velocity

5)maximum height from the ground

1) It takes 5.06 s to the projectile to hit the ground.

2) The range measured from the base of the cliff is 43.8 m.

3) The vertical component of the velocity just before the projectile hits the ground is -44.6 m/s.

4) The magnitude of the velocity when the projectile hits the ground is 45.4 m/s.

5) The maximum height from the ground is 100 m  

Explanation:

Hi there!

The equations for the position and velocity vectors of the projectile are given by the following expressions:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity

v = velocity vector at time t

1) Please, see the attached figure. Notice that when the projectile hits the ground, the y-component of the vector "r final", ry final, is -100 m relative to the launching point. Then, using the equation for the y-component of r we can calculate the time it takes the projectile to hit the ground.

Considering the origin of the frame of reference as the launching point:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = v0 · t · sin α + 1/2 · g · t²

-100 m = 10 m/s · t · sin 30° - 1/2 · 9.8 m/s² · t²

0 = 100 m + 10 m/s · t · sin 30° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 5.06 s

It takes 5.06 s to the projectile to hit the ground.

2) Now, we have to calculate the magnitude of the x-component of "r final", rx final, at time t = 5.06 s.

x = x0 + v0 · t · cos α

x = 0 m + 10 m/s · 5.06 s · cos 30°

x = 43.8 m

The range measured from the base of the cliff is 43.8 m

3) We know the time when the projectile hits the ground, so we can calculate its velocity using the equation of the velocity vector:

v = (v0 · cos α, v0 · sin α + g · t)

We only need the vertical component of the velocity ( v0 · sin α + g · t), however, let´s calculate both because we are going to need the x-component in point 4).

x-component of v = v0 · cos α = 10 m/s · cos 30° = 8.66 m/s

y-component of v = v0 · sin α + g · t

y-component of v = 10 m/s · sin 30° - 9.8 m/s² · 5.06 s

y-component of v = -44.6 m/s

The vertical component of the velocity just before the projectile hits the ground is -44.6 m/s.

(see the figure to better understand why the y-component of v, vy, is negative when the projectile hits the ground).

4) The vector velocity at the time when the projectile hits the ground is:

v = (8.66 m/s, -44.6 m/s)     (the x-component was calculated in point 3).

The magnitude of the velocity is calculated as follows ( notice in the figure that "v" is the hypotenuse of the triangle formed by "v", "vx" and "vy"):

v² = vx² + vy²

v² = (8.66 m/s)² + (-44.6 m/s)²

v = 45.4 m/s

The magnitude of the velocity when the projectile hits the ground id 45.4 m/s.

5) Notice that at the maximum height, the y-component of the velocity vector is 0 ( see vector "v1" in the figure). Then, we can use the equation of the y-component of the velocity vector to obtain the time at which the projectile is at its max-height and with that time we can calculate the y-component of the vector r1 in the figure:

vy = v0 · sin α + g · t

0 = 10 m/s · sin 30° - 9.8 m/s² · t

(-10 m/s · sin 30°) / -9.8 m/s² = t

t = 0.510 s

Now, let´s calculate the y-component of r1:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 0 m + 10 m/s · 0.510 s · sin 30° - 9.8 m/s² · (0.510 s)²

y = 1.02 × 10⁻³ m   ( if you do not round the time obtained in the step before, the value will be 0 m).

The maximum height from the ground is 100 m +  1.02 × 10⁻³ m = 100.0 m

Have a nice day!

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