A shell fired from the ground with an initial speed of 1.70 x 10^3 m/s at an initial angle of 55 degrees to the horizontal. neglecting air resistance, find
a) the shell's horizontal range
b)the amount of time the shell is in motion.
see the other question on the football I answered.
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To find the answers, we can break down the initial velocity of the shell into its horizontal and vertical components.
Given:
Initial speed (v₀) = 1.70 x 10^3 m/s
Initial angle (θ) = 55 degrees
Step 1: Calculate the horizontal and vertical components of velocity.
The horizontal component of velocity (v₀x) can be determined using the equation:
v₀x = v₀ * cos(θ)
Substituting the given values, we find:
v₀x = (1.70 x 10^3 m/s) * cos(55 degrees)
Step 2: Calculate the vertical component of velocity.
The vertical component of velocity (v₀y) can be determined using the equation:
v₀y = v₀ * sin(θ)
Substituting the given values, we find:
v₀y = (1.70 x 10^3 m/s) * sin(55 degrees)
a) Finding the shell's horizontal range:
The horizontal range (R) can be determined by multiplying the horizontal component of velocity (v₀x) by the time of flight (t).
The time of flight (t) can be calculated using the equation:
t = (2 * v₀y) / g
Where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we find:
t = (2 * (1.70 x 10^3 m/s * sin(55 degrees))) / 9.8 m/s²
Once we have the time of flight (t), the horizontal range (R) can be determined using the equation:
R = v₀x * t
Substituting the values, we find:
R = (1.70 x 10^3 m/s * cos(55 degrees)) * t
b) Finding the amount of time the shell is in motion:
Using the same equation as before:
t = (2 * v₀y) / g
Substituting the values, we find:
t = (2 * (1.70 x 10^3 m/s * sin(55 degrees))) / 9.8 m/s²
Simplifying the equation gives us the amount of time the shell is in motion.
So, to summarize:
a) The shell's horizontal range (R) is given by:
R = (1.70 x 10^3 m/s * cos(55 degrees)) * t
b) The amount of time the shell is in motion (t) is given by:
t = (2 * (1.70 x 10^3 m/s * sin(55 degrees))) / 9.8 m/s²
Now, you can substitute the values into these equations to find the answers.