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°.