How long will it take a shell fired from a cliff at an initial velocity of 800m/s at an angle 30 degrees below the horizontal to reach the ground 150m below?
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How long will it take a shell fired from a cliff at an initial velocity of 800m/s at an angle of 30degree below the horizontal to reach the ground 1500m below
To determine the time it will take for a shell to reach the ground, we need to analyze the vertical motion and use the kinematic equations.
First, let's break down the initial velocity into horizontal and vertical components. The horizontal component is given by:
Vx = V * cos(theta)
= 800 * cos(30°)
≈ 800 * 0.866
≈ 692.8 m/s
The vertical component is given by:
Vy = V * sin(theta)
= 800 * sin(30°)
≈ 800 * 0.5
= 400 m/s
We can now use the vertical motion equation to find the time it takes for the shell to reach the ground:
y = yo + Vyi * t - 0.5 * g * t^2
Where:
y = final vertical position (150 m below the cliff)
yo = initial vertical position (0 m)
Vyi = vertical component of initial velocity (400 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation:
-150 = 0 + 400t - 0.5 * 9.8 * t^2
0.5 * 9.8 * t^2 - 400t - 150 = 0
To solve this quadratic equation, we can apply the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 0.5 * 9.8, b = -400, and c = -150.
t = (-(-400) ± sqrt((-400)^2 - 4 * 0.5 * 9.8 * (-150))) / (2 * 0.5 * 9.8)
Simplyfing:
t = (400 ± sqrt(160000 + 2940)) / 9.8
t = (400 ± sqrt(162940)) / 9.8
t ≈ (400 ± 403.66) / 9.8
Using the positive value:
t ≈ (400 + 403.66) / 9.8
t ≈ 803.66 / 9.8
t ≈ 82.05 seconds
Therefore, it will take approximately 82.05 seconds for the shell to reach the ground 150 meters below.
Vo = 800m/s[-30o].
Xo = 800*Cos330 = 693 m/s.
Yo = 800*sin(-30) = -400 m/s.
Y^2 = Yo^2 + 2g*h.
Y^2 = (-400)^2 + 19.6*150 = 162,940
Y = 404 m/s. = Final velocity of ver. component.
Y = Yo + g*Tf.
404 = -400 + 9.8Tf
Tf(Fall time) = ?