# A projectile, fired with unknown initial velocity, lands 18.4 s later on the side of a hill, 2740 m away horizontally and 469 m vertically above its starting point. (Ignore any effects due to air resistance.)

(a) What is the vertical component of its initial velocity?

m/s

(b) What is the horizontal component of its initial velocity?

m/s

(c) What was its maximum height above its launch point?

m

(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?

v = m/s

θ = °

## vertical:

469=Viv*t -1/2 g t^2 solve for viv.

horizontal

2740=vih*t solve for vih

max height: mgh=1/2 m viv^2

solve for h.

vat wall= sqrt(vih^2+vivf^2)

where vivf can be found

1/2 m vivf^2=1/2 mvi^2-mg*496

solver for vivf the final vertical velocity

Now, speed at impace

v= sqrt(vivf^2+vih^2)

angle:

arctan=vivf/vih

## I don't understand the angle part. But thank you so much for the rest of it!

## (a) To calculate the vertical component of the initial velocity, we can use the equations of motion for projectile motion. The vertical motion can be calculated using the equation:

Δy = v₀y * t - (1/2) * g * t²

Where:

Δy is the vertical displacement (469 m),

v₀y is the vertical component of the initial velocity (what we want to find),

t is the time of flight (18.4 s), and

g is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the given values, we can solve for v₀y:

469 = v₀y * 18.4 - (1/2) * 9.8 * (18.4)²

Simplifying the equation:

v₀y * 18.4 = 469 + (1/2) * 9.8 * (18.4)²

v₀y * 18.4 = 469 + 1653.92

v₀y * 18.4 = 2122.92

Dividing both sides by 18.4:

v₀y = 2122.92 / 18.4

v₀y ≈ 115.33 m/s

Therefore, the vertical component of the initial velocity is approximately 115.33 m/s.

(b) To calculate the horizontal component of the initial velocity, we can use the equation:

Δx = v₀x * t

Where:

Δx is the horizontal displacement (2740 m),

v₀x is the horizontal component of the initial velocity (what we want to find), and

t is the time of flight (18.4 s).

Plugging in the given values, we can solve for v₀x:

2740 = v₀x * 18.4

Dividing both sides by 18.4:

v₀x = 2740 / 18.4

v₀x ≈ 149.13 m/s

Therefore, the horizontal component of the initial velocity is approximately 149.13 m/s.

(c) To find the maximum height above its launch point, we can use the equation:

y = v₀y * t - (1/2) * g * t²

We know that at the maximum height, the vertical velocity component (v₀y) will be zero. So we can set v₀y to zero and solve for t:

0 = v₀y - g * t

0 = 115.33 - 9.8 * t

Solving for t:

9.8 * t = 115.33

t ≈ 11.786 s

Now, we can substitute this value of t back into the equation to find the maximum height (y):

y = 115.33 * 11.786 - (1/2) * 9.8 * (11.786)²

Simplifying the equation:

y ≈ 678.6

Therefore, the maximum height above its launch point is approximately 678.6 m.

(d) At any point, the speed of the projectile can be calculated using the equation:

v = sqrt((v₀x)² + (v₀y)²)

Plugging in the values:

v = sqrt((149.13)² + (115.33)²)

v ≈ sqrt(22235.3769 + 13295.5089)

v ≈ sqrt(35530.8858)

v ≈ 188.68 m/s

Therefore, the speed of the projectile as it hit the hill would be approximately 188.68 m/s.

To find the angle (θ) that the velocity makes with the vertical, we can use the equation:

θ = arctan(v₀x / v₀y)

Plugging in the values:

θ = arctan(149.13 / 115.33)

θ ≈ arctan(1.293)

θ ≈ 51.622°

Therefore, the angle that the velocity makes with the vertical is approximately 51.622°.