Answer :
Answer:
θ = 62.72°
Explanation:
The projectile describes a parabolic path:
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = x₀+ vx*t Formula (1)
vx = v₀x
Where:
x: horizontal position in meters (m)
x₀: initial horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
v₀x: Initial speed in x in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)
vfy= v₀y -gt Formula (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 120 m/s , at an angle θ above the horizontal
v₀x= 55 m/s
x-y components of the initial velocity ( v₀)
v₀x = v₀*cosθ Equation (1)
v₀y = v₀*sinθ Equation (2)
Calculating of the angle θ
We replace data in the Equation (1)
55 = 120*cosθ
cosθ = 55 / 120
[tex]\theta = cos^{-1}( \frac{55}{120} )[/tex]
θ = 62.72°
The angle the projectile makes with the horizontal is 63°
To calculate the angle above the horizontal, we use the formula below.
Formula:
- S = Vcos∅............. Equation 1
Where:
- S = Initial horizontal speed of the projectile
- V = Initial velocity of the projectile.
- ∅ = angle above the horizontal
make ∅ the subject of the equation
- ∅ = cos⁻¹(S/V)............... Equation 2
From the question,
Given:
- S = 55 m/s
- V = 120 m/s
Substitute these values into equation 2
- ∅ = cos⁻¹(55/120)
- ∅ = cos⁻¹(0.4583)
- ∅ = 62.72
- ∅ ≈ 63°
Hence, The angle the projectile makes with the horizontal is 63°.
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