Question

A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile's initial horizontal speed is 55 meters per second, then angle θ measures approximately how much?

Answer 1

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

$\theta = cos^{-1}( \frac{55}{120} )$

θ = 62.72°