A shell is fired from the ground with an initial speed of 1.72 103 m/s at an initial angle of 40° to the horizontal.
a) Neglecting air resistance find the horizontal range
b) Find the amount of time the shell is in motion
To find the answers to these questions, we can use the equations of projectile motion. There are two components of motion to consider: the horizontal motion and the vertical motion.
a) To find the horizontal range, we need to determine the distance covered by the shell in the horizontal direction.
First, we can find the initial horizontal velocity (Vx) using the initial speed (Vi) and the angle of projection (θ). This can be calculated using the formula:
Vx = Vi * cos(θ)
Substituting the given values:
Vx = (1.72 * 10^3 m/s) * cos(40°)
Next, we need to find the time of flight (T) for the shell. The time of flight can be determined based on the total vertical displacement (Δy) during the motion. Since we know that the shell will land at the same level as its launch point, the vertical displacement is zero.
The formula for vertical displacement is:
Δy = Vi * sin(θ) * T - (1/2) * g * T^2
Since Δy = 0, we can simplify the equation to:
0 = Vi * sin(θ) * T - (1/2) * g * T^2
We can solve this quadratic equation for T.
Finally, the horizontal range (R) can be calculated by multiplying the time of flight (T) by the horizontal velocity (Vx):
R = Vx * T
b) To find the time of flight, we can rearrange the quadratic equation:
(1/2) * g * T^2 = Vi * sin(θ) * T
Simplifying:
(1/2) * g * T = Vi * sin(θ)
Solving for T:
T = 2 * Vi * sin(θ) / g
Substituting the given values and solving for T will give us the amount of time the shell is in motion.